Sample Chapter - AC Grade 10 v2 - 9. Trigonometry

Trigonometry 9

Big ideas

• Right triangles have special properties that allow their side lengths to be determined using the Pythagoras’ theorem or ratios.

• Trigonometry connects the angle measures and side lengths of a triangle and provides a set of functions that can be used to find unknown angle measures and side lengths.

Curriculum Outcomes

AC9M10M03

Solve practical problems applying Pythagoras’ theorem and trigonometry of rightangled triangles, including problems involving direction and angles of elevation and depression

Chapter outline

9.01 Pythagoras’ theorem

9.02 Pythagoras in 3D

9.03 Trigonometric ratios

9.04 Trigonometry in 3D

9.05 Angles of elevation and depression

9.06 Bearings

9.07

9.01 Pythagoras’ theorem

After this lesson, you will be able to…

• apply Pythagoras’ theorem to identify right-angled triangles.

• use Pythagoras’ theorem to find the length of the hypotenuse and other sides in a right-angled triangle.

• apply knowledge of Pythagoras’ theorem to solve real-world problems.

Pythagoras’ theorem

Hypotenuse

Right angle

In a right-angled triangle, the largest angle in the triangle is 90°. The side opposite the right-angle will be the longest side. We call this side the hypotenuse

c

All three sides of a right-angled triangle are related by the equation shown.

The two smaller sides will be called a and b, and the hypotenuse (the longest side) will be c. a 2 + b2 = c 2

a, b, c Sides of a right-angled triangle

a, b The lengths of the shorter sides

c The length of the hypotenuse

This relationship between sides in a right-angled triangle is called Pythagoras’ theorem. We can use this theorem to:

• Check whether or not a triangle is a right-angled triangle

• Find the length of the hypotenuse

• Find the length of one of the shorter sides in a right-angled triangle

Interactive exploration

Explore online to answer the questions

mathspace.co

Use the interactive exploration in 9.01 to answer these questions.

1. What do you notice about the sum of the area of the squares of the two shorter sides and the area of the square of the hypotenuse?

2. If you subtracted the area of the shorter side sides from the largest side, what do you get?

3. Can you find another Pythagorean triple, where all three sides have integer values?

Example 1

Use Pythagoras’ theorem to determine whether this is a right-angled triangle.

a Let a and b represent the two shorter side lengths. First find the value of a 2 + b2 .

Create a strategy

Substitute the lengths of the two shorter sides and add them.

Apply the idea

a 2 + b2 = (9)2 + (16)2

Substitute the value of a and b = 81 + 256 Evaluate = 337 Evaluate the sum

b Let c represent the length of the longest side. Find the value of c 2

Create a strategy

Substitute the length of the longest side.

c Is the triangle a right-angled triangle?

Create a strategy

Compare the results from parts (a) and (b) to determine if a 2 + b2 = c 2

Apply the idea c 2 = (18)2

Substitute the value of c = 324 Evaluate

Apply the idea 337 ≠ 324

The triangle is not a right-angled triangle because a 2 + b2 ≠ c 2 .

Apply the idea

c 2 = a 2 + b2

c 2 = 122 + 162

c 2 = 144 + 256

c 2 = 400

Write the formula

Substitute a and b

Evaluate the squares

Evaluate the sum

c = Take the square root of both sides

c = 20

Evaluate the square root

6 x

Create a strategy

Use Pythagoras’ theorem: a 2 + b2 = c 2

Apply the idea

We can substitute the following a = x, b = 6, and c = 10 into the formula:

a 2 + b2 = c 2

x 2 + 62 = 102

x 2 + 36 = 100

x 2 + 36   36 = 100 – 36

x 2 = 64

Write the formula

Substitute the values

Evaluate the squares

Subtract 36 from both sides

Evaluate the difference

x = Take the square root of both sides

x = 8

Reflect and check

Evaluate the square root

The most important thing to remember when finding a short side is that the two lengths need to go into different parts of the formula.

If we substitute incorrectly, we will end up with the square root of a negative number (calculator error) or a side which is longer than the hypotenuse.

Example 4

A movie director wants to shoot a scene where the hero of the film fires a grappling hook from the roof of one building to the roof of another.

If the first building is 37 m tall, the other building is 54 m tall and the street between them is 10 m wide, what is the minimum length, l, of rope needed for the grappling hook? Round your answer to two decimal places.

Create a strategy

Use the Pythagoras’ theorem: c 2 = a 2 + b2 .

Apply the idea

Notice that we have a right-angled triangle, and the length of the grappling hook’s rope will be the hypotenuse. The one short side is the width of the street, 10, and the other short side is the difference in heights between the two buildings, 54   37.

We can substitute the following a = 10, b = (54   37), and c = l into the formula:

c 2 = a 2 + b2

l2 = 102 + (54   37)2

l2 = 102 + 172

l2 = 100 + 289

l2 = 389

Write the formula

Substitute the values

Simplify the brackets

Evaluate the indices

Evaluate the addition

l = Take the square root of both sides

l ≈ 19.72

Evaluate the square root

Idea summary

a b c

To find the hypotenuse:

To find a shorter side use:

Pythagoras’ theorem relates the three sides of a rightangled triangle, a and b are the two smaller sides, and the longest side, called the hypotenuse, is c

We can also test to see if a triangle is right- angled by checking to see if its three sides satisfy a 2 + b2 = c 2

2 = a 2 + b2

We can take the square root of both sides to give us the following formulas:

Practice questions

Choose your question path

Explorer

1–3, 4ace, 5a-dg, 7ac, 9, 10, 12ab i, ii, 13–15, 17

Understanding

Adventurer

2cd, 3, 4bdf, 5deh, 6, 7d, 8a, 9, 10, 12ab ii, iii, 13–17

1 Write Pythagoras’ theorem in a form you could use to find:

a The length of c

b The length of a.

Trailblazer

2d, 3, 4gh, 5dfi, 6, 8, 9, 11, 12, 13b, 15, 16, 18

2 Calculate the value of x, rounded to two decimal places. All values of x are positive. a b x 2 = 22 + 32 c 82 = 52 + x 2 d

3 Use Pythagoras’ theorem to determine whether this is a right-angled triangle.

a Let a and b represent the two shorter side lengths. First find the value of a 2 + b2 .

b Let c represent the length of the longest side. Find the value of c 2 .

c Is the triangle a right-angled triangle?

Fluency

4 Are these Pythagorean triples?

a (12, 16, 26) b (9, 12, 15)

c (4, 8, 12) d (12, 16, 20)

e (8, 15, 17) f (7, 7, 10) g (8, 11, 14) h (7, 24, 25)

5 Find the length of the unknown side as an exact value: a

g A right-angled triangle has a one side of length 6 and another side of length 4. Find the length of the hypotenuse.

h A right-angled triangle has a hypotenuse of length 9 and another side of length 3. Find the length of the third side.

i A right-angled triangle has a hypotenuse of length 17 and another side of length 14. Find the length of the third side.

6 Two poles, one of length 8 m, and the other of length 14 m, stand 8 m apart on level ground. Find the length of wire required to join the tops of the poles.

7 Find the length of a diagonal of each shape, rounded to one decimal place where appropriate.

8 Find the distance between the points P and Q, rounded to one decimal place.

9 Two buildings, one of height 300 m and the other of height 450 m, stand a distance apart on level ground. If a cable joining the tops of the buildings is 200 m long, find, to the nearest metre, the distance between the buildings.

Reasoning

10 Can a triangle with sides of length 3 cm, 5 cm and 7 cm be a right-angled triangle? Use Pythagoras’ theorem to justify your answer.

11 If AB = 10 cm, AD = 8 cm and ∠ADB is a right angle, show that D is the midpoint of BC

12 A triangle with sides 3, 4, and 5 units is a right triangle.

a If the triangle is enlarged by a factor of 3, does Pythagora’s theorem still hold? Justify your answer.

b Is a triangle with these sides a right triangle? i 30, 40, 50 ii 35, 45, 55 iii 3

Problem-solving

13 Find the value of x, rounded to one decimal place. a b

14 a Find the length of the diameter AB, of the circle.

b Find the exact length x.

15 Casper produces thin plastic triangles.

a If the triangles need to be edged by a material that costs 3 cents per cm, how much, in dollars, will it cost to edge 100 triangles?

b If the triangles need to be painted with a coating that costs 2 cents per cm2, how much, in dollars, will it cost to coat 100 triangles.

16 The diagram shows four squares. The outer square has side lengths of 20 cm. The next square is formed by joining the midpoints of the outer square, the third square by joining the midpoints of the second square, and the fourth square by joining the midpoints of the third square. Tamsin is using wire to construct the shape in this diagram. Find, to the nearest centimetre, the length of wire she requires.

17 Tricia’s house has the outer dimensions as shown in the diagram: A drainpipe needs to run from the vertex of the roof, indicated by point A in the diagram, vertically downwards until it is 30 cm above the ground. Find the length of pipe needed in centimetres, rounded to two decimal places.

18 Angelia and Beth start their hikes from the same starting point. Angelia walks 900 m south and then 900 m east, before stopping for a lunch break. Beth stops for a break having walked 1200 m south and then 600 m east. If they both walked back to the start point in a straight line, who would walk furthest, and by how far to the nearest metre?

9.02

Pythagoras in 3D

After this lesson, you will be able to…

• recognise and identify right-angled triangles in different 3D shapes like cubes, cones and prisms, and relate them to 2D slices of those shapes.

• calculate the lengths of sides and diagonals in 3D objects using right-angled triangles and Pythagoras’ theorem.

• apply Pythagoras’ theorem to real-world problems.

Pythagoras in 3D

We can identify right-angled triangles in 3D space the same way that we do in 2D space, by looking for a right angle.

In 3D space, right angles occur between lines that are perpendicular in the same 2D plane. In other words, if two lines meet at a right-angle in some 2D slice, the triangle formed with these two lines is a right-angled triangle.

We can look at 2D slices of this cube

If we take a slice through the vertices A, B, C and D we can see that any triangle determined by three of these vertices will be right-angled, since this slice gives us a square.

In fact, any of the 2D slices that corresponds to a face of the cube will give us multiple possible right-angled triangles.

Taking a 2D slice through the vertices F, H, B and D gives us a rectangle, which means that we still form a right-angled triangle from any three of these vertices.

E If we return to 3D space, we can see how these right-angled triangles look in 3D.

We can perform a similar exercise with other 3D solids like cones and prisms to find other rightangled triangles in 3D space:

In most straight-edged solids, the diagonals formed by joining two non-adjacent vertices will often be the hypotenuse of some right-angled triangle in that solid.

Interactive exploration

Explore online to answer the questions

mathspace.co

Use the interactive exploration in 9.02 to answer these questions.

1. How many right-angled triangles in 3D can you find in a rectangular prism?

2. For each triangle, what do you notice about where the right angle is?

3. For a rectangular prism with whole number side lengths, why do you think some of rightangled triangle side lengths are not whole numbers?

We can find lengths in 3D using these steps as guidance:

1. For the desired length, draw a 2D right-angled triangle involving that length using as many vertices as possible.

2. Find the other two lengths of the 2D right-angled triangle. This may involve drawing another triangle and Pythagoras’ theorem.

3. Use Pythagoras’ theorem to find the desired length.

Example 1

A square prism has sides of length 9 cm, 9 cm and 16 cm as shown.

a Calculate the length of HF, leaving your answer in exact form.

Create a strategy

We can use Pythagoras’ theorem to find the length of HF in the triangle △FGH.

Apply the idea F G H E z cm 9 cm 16 cm

We can start by drawing the 2D slice and labelling the rightangled triangle we can use to find the length of HF.

HF2 = HG2 + GF2

z 2 = 92 + 162

z 2 = 337

Use Pythagoras’ theorem

Substitute the lengths

Evaluate the powers and addition

z = Take the square root of both sides

So the value of HF is  .

b Find the length of the diagonal DF, rounded to two decimal places.

Create a strategy

We can use Pythagoras’ theorem to find the length of DF in the triangle △DHF.

Apply the idea

We can now draw the 2D slice which includes DF and label the right-angled triangle.

Use Pythagoras’ theorem

Substitute the lengths

Evaluate the squares

Evaluate the addition

Take the square root of both sides

Evaluate and round

So the value of DF is 20.45 cm.

Reflect and check

If we were just asked to find DF without first being asked to find HF, would you have used this same strategy or something different?

There are many ways to find DF, but in all cases you will need to solve for one other length that isn’t given on the diagram.

Example 2

A soft drink can has a height of 13 cm and a radius of 5 cm. Find L, the length of the longest straw that can fit into the can (so that the straw is not bent and fits entirely inside the can).

Round your answer down to the nearest centimetre, to ensure it fits inside the can.

Create a strategy

Use the Pythagoras’ theorem to find the length L using the height and radius of the can.

Apply the idea

10 cm 13 cm L

c 2 = a 2 + b2

L2 = 132 + 102

L2 = 269

Making a 2D slice, L is the hypotenuse of the right-angled triangle. The other sides are the height which is 13 and the diameter of the can which is 2 × 5 = 10.

Use Pythagoras’ theorem

Substitute the height, diameter and L

Evaluate the powers and addition

L = Take the square root of both sides

= 16

Evaluate and round

The length of the longest straw to the nearest centimetre is 16 cm.

Idea

summary

We can find right-angled triangles in a 3D solid by taking 2D slices. Once we identify rightangled triangles, we can use Pythagoras’ theorem to find unknown lengths in the solids.

Practice questions

Choose your question path

Explorer

1, 3–5, 7, 9, 11, 13, 14, 17–19

Understanding

Adventurer

2–5, 7, 9, 12–14, 17–20

Trailblazer

2–4, 6–8, 10, 12–15, 19–23

1 Can Pythagoras’ theorem be applied to any triangle? If not, specify the type of triangle for which it can be applied.

2 The diagram shows a rectangular prism with side lengths a, b and c. We want to find the length of the diagonal from B to H:

a Do we know all of the lengths in △BFH? If not, which length can we work out from the given information.

b What triangle could we use to work out the length x?

c Once we find x, can we then work out length y?

d Write an expression for y in terms of c and x.

Fluency

3 All edges of the cube are 5 cm long. x is the diagonal of the base and y is the diagonal through the centre of the cube. Find x and y, rounded to two decimal places.

4 A square-based prism has dimensions of 12 cm by 12 cm by 15 cm. Calculate, with answers rounded to two decimal places:

a The length of HF.

b The length of DF

5 A rectangular prism has dimensions as labelled on the diagram. Find each length in exact form:

a EG

b AG

c DG

6 Consider the triangular prism: Find, rounded to two decimal places:

a The value of EC.

b The value of XC

c The value of DF

7 Consider the rectangular prism. Find, rounded to two decimal places:

a The value of EG.

b The value of AG

A new diagonal DG has been added.

c Find the length of DG to two decimal places.

d Calculate the area of the right-angled triangle △ADG.

8 This triangular prism shaped box labelled ABCDEF needs a diagonal support inserted as shown:

a Write an expression for the length of BF in terms of BD and DF.

b Find the length of AF in terms of AB, BD and DF

c If AB = 19, BD = 30 and DF = 43, find the length of AF, rounded to two decimal places.

d Find the length of AF if AB, BD and DF increase by 10. Round your answer to two decimal places.

9 A triangular prism has dimensions as shown. Find the exact length of:

a CE

b BE

c CX

10 The square pyramid has its apex, V, aligned directly above the centre of its base, W

a Calculate the length of AW, rounded to two decimal places.

b Find the length of VW, rounded to two decimal places.

11 Consider the cone with slant height of 13 m and perpendicular height of 12 m:

a Find the length of the radius, r.

b Find the diameter of the cone’s base.

12 A right-angled triangular divider has been placed inside a box, as shown:

a Calculate the area of the triangular divider, rounded to two decimal places.

b Calculate the perimeter of the triangular divider, rounded to two decimal places.

13 A soft drink can has a height of 11 cm and a radius of 4 cm. Find L, the length of the longest straw that can fit into the can.

Round your answer down to the nearest cm, to ensure it fits inside the can.

Reasoning

14 Using the diagram, explain how to show that the length of the diagonal AG is . Name all triangles used.

15 The diagram shows a rectangular-based pyramid ABCDE, F is the centre of the base. Suppose that M is the midpoint of AD and that N is the midpoint of CD

If we were going to calculate the surface area of the pyramid, state the additional lengths we would need to calculate, giving them in terms of a, b and c. Name the triangles that would be needed for the calculations.

16 In the diagram, the length of AG is units.

Given that a ≥ b ≥ c, find a set of integer values of a, b and c that will give the correct length for AG. Is this the only set that works? Explain.

Problem-solving

17 A rectangular prism shaped garden shed has a height of 2.4 m, length of 0.94 m and width of 4.2 m.

a Sketch a diagram of a wooden plank that is placed at the greatest possible length inside the garden shed.

b Find the length of the plank of wood, rounded to one decimal place.

18 A cylindrical can of drink has a circumference of 7π cm and a height of 12 cm. Find the length of the longest straw that can fit into the can. Round your answer down to the nearest cm, to ensure it fits inside the can.

19 A juice container has the shape of a rectangular prism. It needs a straw that must extend 20 mm beyond the container while touching the furthest corner of the base.

Calculate the length of the straw needed. Round your answer to the nearest millimetre.

20 A builder needs to carry lengths of timber along a corridor in order to get them to where he is working. There is a rightangled bend in the corridor along the way. The corridor is 2 m wide and the ceiling is 2.5 m above the floor:

a Calculate the maximum length of timber, in exact form, that would fit in the corridor when held parallel to the ground.

b Since the corridor has a height of 2.5 m, he can fit a longer piece of timber around the corner by angling it so that it reaches from the floor to the ceiling.

Find the maximum length of timber that is able to fit through the corridor, rounded to two decimal places.

21 Consider the net of a pyramid with a rectangular base:

• APQRC is a straight line

• PA = 12 cm, RC = 9 cm, HG = 15 cm, and EH = 8 cm

• EA = HA and FC = GC

• When the net is folded, points A, B, C, and D all meet at S, which lies vertically above the point Q along PR

a Show that △PSR is a right triangle.

b Find the area of △PSR

c Find the height of the pyramid SQ

d Using the formula , find the volume of the pyramid.

22 Consider the rectangular prism aquarium with dimensions of 25 cm by 30 cm by 48 cm.

The top has points JKLM while the base has points PQRS, respectively. A tree branch is resting at points J and T where T is 40 cm away from Q

a Find the length of PT to the nearest tenth.

b Find the length of the tree branch JT to the nearest tenth.

c Find the length ST to the nearest tenth.

d A fish at point L swims to point P. Find the distance, LP, that the fish travelled, to the nearest tenth.

23 Find the smallest integer value of h, so that a straw of length 12 cm can fit into the box shown:

9.03 Trigonometric ratios

After this lesson, you will be able to…

• identify and define sine, cosine, and tangent, and describe their relationship between angles and sides in right-angled triangles.

• use trigonometric ratios to find an unknown side or angle in right-angled triangles.

Trigonometric ratios

In right-angled triangles, we can use trigonometric ratios to describe the relationship between a given angle and two sides.

The trigonometric ratios are:

Hypotenuse

Exploration

We have three right-angled triangles with acute angles of 30° and 60°.

1. Is △ABC ∼ △DEF ∼ △GHI? Explain why or why not.

2. Find the ratio for sin 30° for each of the triangles.

3. What do you notice about each ratio (use a calculator if you need to)?

4. Find the ratio for cos 30° for each of the triangles.

5. What do you notice about each ratio (use a calculator if you need to)?

6. Find the ratio for tan 30° for each of the triangles.

7. What do you notice about each ratio (use a calculator if you need to)?

8. Draw another similar triangle which would have the same ratios.

The trigonometric ratios will be the same for similar right-angled triangles. This is because the size of the angles in a triangle determines the ratios between the side lengths of a triangle.

Example 1

Evaluate cos (θ) within △ABC

Create a strategy

Identify which sides we need for the cosine ratio and then fill in the side lengths. B

Apply the idea

Hypotenuse

The angle θ is also called ∠B. We are using the cosine ratio

so we need the lengths of the adjacent and hypotenuse sides.

Adjacent

With respect to the angle θ, the adjacent side is 5, and the hypotenuse is the longest side of 13.

Defintion of cosine

Substitute the sides

Idea summary

The trigonometric ratios are:

Hypotenuse

Adjacent Opposite

Find a side

If we are given one side length and one angle, we can use the trigonometric ratios to find one of the missing side lengths.

We can follow these steps to find a missing side in a right-angled triangle:

1. Highlight the reference angle.

2. Identify which sides are the hypotenuse, opposite, and adjacent — label if desired.

3. Choose the ratio which has one of the required sides given and one is the missing side:

• Sine: opposite and hypotenuse

• Cosine: adjacent and hypotenuse

• Tangent: opposite and adjacent

4. Set up the chosen ratio:

5. Solve for the unknown.

Remember the following:

• For a given angle θ, the value of each trigonometric ratio stays the same no matter the size of the triangle.

• Your calculator must be set to degree mode when working with degrees in trigonometric ratios.

Example 2

Find the value of the variable, rounded to two decimal places: a

Create a strategy

Identify the given sides and write an equation using the appropriate trigonometric ratio.

Apply the idea

With respect to the angle 47°, the length of the adjacent side is f, and the length of the hypotenuse is 8, so we can use the cosine ratio.

Use the cosine ratio

Substitute the values and f Multiply both sides by 8 Evaluate and round

Create a strategy

Use the trigonometric ratio of sine and substitute the given values.

Apply the idea

With respect to the angle 42°, the length of the opposite side is 11, and the length of the hypotenuse is g, so we can use the sine ratio.

Use the sine ratio

Substitute the values and g

Multiply both sides by g

Divide both sides by sin 42°

Evaluate and round

Reflect and check

When the unknown is in the denominator, we need to take care to rearrange properly.

Idea summary

We can use the trigonometric ratios to find an unknown side length of a right-angled triangle.

Once we set up our equation with a variable representing the unknown side length, we can use inverse operations to make the variable the subject of the equation.

Then we can evaluate the expression in our calculators to find the side length.

Find an angle

If we are given two sides of a right-angled triangle, we can use the trigonometric ratios to find a missing angle.

We can follow these steps:

1. Using the unknown angle as the reference angle, label the given sides as opposite, adjacent, or hypotenuse.

2. Fill in the trigonometric ratio with the given information.

3. Use the inverse ratio to solve for the angle. Example 3 Find the value of x to the nearest degree.

Create a strategy

Use the cosine ratio to set up the equation then use the inverse ratio to find the angle.

Apply the idea

With respect to the angle of x, the length of the adjacent side is 7, and the length of the hypotenuse side is 25, so we can use the cosine ratio.

Use the cosine ratio

Substitute the values

Apply the inverse ratio

Evaluate and round

Reflect and check

If we get an error on our calculator, it usually means we haven’t used the right ratio or substituted incorrectly.

Idea summary

To find an unknown angle of a right-angled triangle, we use a trigonometric ratio to set up an equation relating the angle and two given sides. Then we can use the inverse ratio to find the value of the angle.

Multi-step problems

In some cases, a single application of trigonometry, on a single right-angled triangle, will not be enough to solve to the question.

In this example, to find the length of h, we must first use our knowledge of trigonometry, in right-angled triangles, to find the length of x. Once we find the length of x, we add 70 to x, to get the length of h.

The vertical distance, h, from the ground to the kite is the sum of the distance from the ground to where the string starts and the vertical distance, x, from the string to the kite. That means h = 2.1 + x, where x = 21.85. So, h = 2.1 + 21.85 = 23.95. The kite is 23.95 m from the ground.

Example 5

Find the value of the variables x and y, rounded to the nearest whole number.

Create a strategy

Use the tangent ratio to find y, then use the cosine ratio to determine x

Apply the idea

With respect to the angle of 64, the length of the opposite side is 8, and the length of the adjacent side is y, so we can use the tangent ratio.

Use the tangent ratio

Substitute the values

Multiply both sides by y

Divide both sides by tan 64°

Evaluate and round

With respect to the angle of 64, the length of the adjacent side is 8, and the length of the hypotenuse is x, so we can use the cosine ratio.

Use the cosine ratio

Substitute the values

Multiply both sides by x

Divide both sides by cos 64°

Evaluate and round

Idea summary

In multi-step problems, a single application of trigonometry will not be enough to solve the problem. Extra addition, subtraction or another application of trigonometry may also be needed.

Practice questions

Choose your question path

Explorer Adventurer

1, 2, 3ace, 4, 5a, 6–8, 10–12, 14, 16, 17abc, 18, 20, 22

3df, 4, 5b, 7, 8, 10, 11a, 12, 14, 16, 17bcd, 18, 20, 22, 23

Trailblazer

3d, 4, 5b, 7, 9, 10, 11b, 13, 15, 16, 17cde, 19, 21–24

Understanding

1 Name the hypotenuse. a b

2 a Which side is opposite to angle θ?

b Which side is adjacent to angle θ?

c Which side is opposite angle α?

d Which side is adjacent to angle α?

e Which angle is opposite the hypotenuse?

3 With reference to the angle θ, find the ratio:

4 Write down the indicated ratios for each triangle using the side labels:

5 Write down the indicated ratios for each triangle:

6 Find the value of tan θ.

7 Find the value of the variable, rounded to two decimal places:

Reasoning

18 An isosceles triangle has equal side lengths of 10 cm and a base of 8 cm as shown.

Explain how to calculate the size of angle A, rounded to one decimal place.

19 If tan θ = 0.4, explain how to find the value of b

Problem-solving

20 If , find the value of cos θ and tan θ, without calculating the size of angle θ

21 If , find the value of cos θ and sin θ, without calculating the size of angle θ

22 Consider the octagon:

a Use trigonometry to calculate the size of each internal angle rounded to the nearest whole number.

b Confirm your answer to part (a) using the formula to find an internal angle of a polygon with n sides:

23 A sporting ground is in the shape of an irregular hexagon.

a Find the perimeter of the ground.

b Find the size of each of the four corner angles of the ground.

24 Calculate the size of angle DEC 9.04

Trigonometry in 3D

After this lesson, you will be able to… • apply trigonometric ratios in 3D.

Trigonometry in 3D

The key difference between 3D and 2D space when using trigonometry is that there are a lot more triangles we can find. Using this abundance of triangles, we can use multiple applications of trigonometry to find previously unknown side lengths and angles in a given problem.

Example 1

Consider the given rectangular prism:

a Find the length x

Create a strategy

Use Pythagoras’ theorem c 2 = a 2 + b2

Apply the idea

The lengths x, 9 and 12 create a right-angled triangle on the base of the prism where x is the hypotenuse.

x 2 = 92 + 122 Use Pythagoras’ theorem

x 2 = 225 Evaluate

x = 15 cm Take the square root of both sides

b Find the length of the prism’s diagonal y

Create a strategy

Use Pythagoras’ theorem c 2 = a 2 + b2

Apply the idea

The lengths x = 15, 8 and y create a right-angled triangle where y is the hypotenuse.

y 2 = 82 + 152 Use Pythagoras’ theorem

y 2 = 289 Evaluate

y = 17 cm Take the square root of both sides

c Find the angle θ to the nearest degree.

Create a strategy

Use a trigonometric ratio.

Apply the idea

In the triangle with angle θ, x = 15 is the adjacent side, and y = 17 is the hypotenuse. So we can use the cosine ratio to find θ

Use the cosine ratio

Find the inverse cosine of both sides

Evaluate and round

Example 2

Asad and Billie are trying to find the height of a pole, but can’t reach the top of the pole. Asad is standing at point A and is 100 m from Billie at point B. On the diagram ∠BAC = 30° and ∠DBC = 48°.

a Find BC, rounded to two decimal places.

Create a strategy

We can use △ABC to find BC. Since we know one side and one angle and it is a right-angled triangle we can use a trigonometric ratio.

Apply the idea

Since we have the adjacent side and are looking for the opposite side we will use the tangent ratio.

The length of BC is about 57.74 m.

Reflect and check

If we use BC, there will be some error introduced as it is a rounded value.

C D h m Pole

b Find the height of the pole.

Create a strategy

We can use △BCD as it is a right-angled triangle and we know one side and one angle.

Apply the idea

Since we have the adjacent side and are looking for the opposite side we will use the tangent ratio.

Height of the pole is about 64.13 m.

Idea summary

To solve trigonometric problems in 3D we can use any of the tools that we have previously learnt:

• Pythagoras’ theorem

• Trigonometric ratios

Practice questions

Choose your question path

1–5, 8, 10, 11, 14, 15, 17 2–8, 10, 13–15, 17, 18 3, 5, 6, 8, 9, 11, 12, 14–16, 18, 19

Understanding

1 A straw with length L has slipped inside a cylindrical can. Use Pythagoras’ theorem to write a rule which could be used to find L.

2 This box shaped like a triangular prism needs a diagonal support inserted between A and F as shown.

Write an expression for the length of BF in terms of BD and DF.

3 All edges of the cube are 5 cm long.

a Draw a labelled diagram of the triangle on the base of the cube which has x as its hypotenuse.

b Find the value of x, rounded to two decimal places.

c Draw a labelled diagram of the triangle through the middle of the cube which has y as its hypotenuse.

d Find the value of y, rounded to two decimal places.

Fluency

4 All edges of the cube are 7 cm long:

a Find the length of these sides, rounded to two decimal places: i EG ii AG

b Find these angles to the nearest degree: i ∠EGH ii ∠EGA iii ∠AHG iv ∠AGH

5 A square prism has sides of length 3 cm, 3 cm and 14 cm.

a Find z, the length of diagonal HF rounded to two decimal places.

b Find θ, rounded to two decimal places.

6 A rectangular box has a triangular divider placed inside it.

a If z = AC, find the value of z, rounded to two decimal places.

b Find the area of the divider, rounded to two decimal places.

7 Consider the triangular prism. Round all answers to two decimal places.

a Find the size of ∠AED.

b Find the length of CE

c Find the length of BE.

d Find the size of ∠BEC

e Find the length of CX

f Find the size of ∠BXC

8 Consider the rectangular prism:

a Find the length x

b Find the length of the prism’s diagonal y, rounded to two decimal places.

c Find angle θ to the nearest degree.

9 Consider the cube:

a Find the size of: i

b Calculate the length of x, rounded to two decimal places.

c Calculate the length of y, rounded to two decimal places.

d Find the angle θ to the nearest degree.

10 A 25 cm × 11 cm × 8 cm cardboard box contains an insert (the shaded area) made of foam:

a Find the length of b, the base of the foam insert, rounded to two decimal places.

b Find the area of foam in the insert, rounded to the nearest square centimetre.

c Find the value of θ rounded to the nearest degree.

11 A, C, and X are three points in a horizontal plane and B is a point vertically above X as shown:

Find these lengths to the nearest metre: a BC b XB

12 Consider the right pyramid with a square base with side length 12 cm. The edge length VA is 23 cm.

a If z = AW, calculate the value of z.

b If θ = ∠VAW, find the value of θ

13 The pyramid has a square base with side length 8 cm and all slanted edges are 12 cm in length. Round all answers to two decimal places.

a Find the length of MD.

b Find the size of ∠PDM.

c Find the height of the pyramid, MP.

d Find the length of MN.

e Find the length of PN.

f Find the size of ∠PNM.

g Find the size of ∠PDC.

14 Antonio and Blake are trying to find the height of a pole, but can’t reach the top of the pole. Antonio is standing at point A and is 80 m from Blake at point B. On the diagram

∠BAC = 27° and ∠DBC = 55°.

a How far is Blake from the bottom of the pole, rounded to two decimal places?

b How tall is the pole, rounded to two decimal places?

Reasoning

15 A cone has radius 7 cm and a slant height of 13 cm.

a Explain using a diagram how it is possible to find angle θ, at the top of the cone, when it is not in a right-angled triangle.

b Find the size of θ, rounded to the nearest degree.

16 A pyramid has a square base with side length 14 m and a vertical height of 24 m.

Find the length of the edge AE using two different methods.

Problem-solving

17 A tree stands at the corner of a square playing field. Each side of the square is 130 m long. At the centre of the field the top of the tree is at an angle of 22° as shown:

a Find the distance from the tree to the centre of the field. Round your answer to two decimal places.

b Find the height of the tree. Round your answer to two decimal places.

c Find the size of the angle from the corner of the field opposite the tree to the top of the tree. Round your answer to the nearest degree.

d Find the size of the angle from an adjacent corner of the field to the top of the tree. Round your answer to the nearest degree.

18 In the tetrahedron shown, the angles ∠VBC, ∠VBA, and ∠ABC are all right-angles.

a Find these distances, rounded to two decimal places: i VA ii VC iii AC

b Find the size of angle ∠VCA. Round your answer to the nearest degree.

Explain how you found your answer.

19 A cone has diameter 16x and a slant height of 17x. Find the perpendicular height of the cone.

9.05 Angles of elevation and depression

After this lesson, you will be able to…

• define angles of elevation and depression and understand their real-world applications.

• model the relative position of two objects using a right-angled triangle.

• identify and apply the three distances (horizontal, vertical, and direct) between two objects when given the angle of elevation or depression.

• solve problems involving angles of elevation and depression using trigonometric ratios.

Angles of elevation and depression

An angle of elevation is the angle from the lower object to the higher one. B

The angle of elevation from point A to B is the angle between the horizontal line at A and the line connecting the two points.

An angle of depression is the angle from the higher object to the lower one. B

The angle of depression from point B to A is the angle between the horizontal line at B and the line connecting the two points.

Both angles are measured with respect to the horizontal.

Exploration

Rectangle ABCD has a diagonal drawn on it.

1. What is the value of x? What property did you use to find this? Is there another way to confirm this?

2. What is the angle of elevation from point C?

3. What is the angle of depression from point A?

4. What do you notice about the angle of elevation and depression?

Notice that the angle of elevation between two points will always be equal to the angle of depression between those two points. This is because they are alternate angles on parallel lines (since all horizontal planes are parallel to each other).

When given the angle of elevation or depression between two objects, we will always be able to model their relative position using a right-angled triangle.

There are three distances between two objects: horizontal distance, vertical distance and direct distance. These will represent the adjacent, opposite and hypotenuse sides respectively. If any two of these are given, then we can find the angle of elevation or depression.

Find the angle of depression from point B to point D. Use x as the angle of depression and round your answer to two decimal places.

Create a strategy

Identify where the angle of depression is and apply the appropriate trigonometric ratio.

Example 1

Apply the idea

The angle of depression is the angle ∠CBD, with adjacent side BC = 9 and hypotenuse BD = 19. We can let x be the unknown angle.

Use the cosine ratio

Substitute the values

Apply inverse cosine on both sides

Evaluate using a calculator and round

Example 2

A fighter jet, flying at an altitude of 4000 m is approaching a target. At a particular time the pilot measures the angle of depression to the target to be 13°. After a minute, the pilot measures the angle of depression again and finds it to be 16°.

a Find the distance AC, to the nearest metre.

Create a strategy

Use the trigonometric ratio of tangent.

Apply the idea

With respect to the given angle 13°, the opposite side is 4000, and the adjacent side is AC, so we can use the tangent ratio.

Use the tangent ratio

Substitute the values

Multiply both sides by AC

Divide both sides by tan 13°

Evaluate using a calculator and round

B

b Find the distance BC, to the nearest metre.

Create a strategy

Use the same method that we used to find AC in part (a).

Apply the idea

With respect to the given angle 16°, the opposite side is 4000, and the adjacent side is BC, so we can use the tangent ratio.

Use the tangent ratio

Substitute the values

Multiply both sides by BC

Divide both sides by tan 16°

Evaluate using a calculator and round

c Now find the distance covered by the jet in one minute to the nearest metre.

Create a strategy

Find the difference between the length of AC and BC

Apply the idea

The distance covered by the jet in one minute is given by the length of AB

AB = 17 326   13 950 Subtract BC from AC = 3376 m Simplify the difference

A room is rectangular and measures 5 metres in length and 4 metres in width. The angle of elevation from the bottom left corner to the top right corner of the room is 57°.

a Draw a diagram of the room.

Create a strategy

To draw a diagram of the room, we need to keep in mind that the room is rectangular in shape and the given dimensions are in metres. Additionally, the angle of elevation from one corner of the room to the other is given, which means we are considering a 3D model.

Apply the idea

Example 3

b Find d, the distance from one corner of the floor to the opposite corner of the floor. Leave your answer in surd form.

Create a strategy

Apply the Pythagora’s theorem, c 2 = a 2 + b2 .

Apply the idea

d2 = 52 + 42

d2 = 41

Substitute the values

Evaluate the right-hand side

d = Take the positive square root of both sides

c Find h, the height of the room. Give your answer to two decimal places.

Create a strategy

Remember the mnemonic SOH-CAH-TOA. Set up an equation involving the height, h, the diagonal on the floor, , and the angle of elevation, 57°.

Apply the idea

Substitute the values

Multiply both sides by Evaluate using a calculator and round

Idea summary

We can solve problems involving angles of elevation and depression using these steps:

1. Draw a diagram if not given one.

2. Label all known information on the digram, including the right angle.

3. Identify which angles or sides need to be found and in what order.

4. Set up the correct trigonometric ratio and solve for the missing side or angle.

5. Give your answer as a sentence.

Practice questions

Choose your question path

1–7, 10–13, 15, 17–21

Angle of

2, 3, 5–13, 15, 18–21, 23 3, 5, 8–11, 13, 14, 16, 20, 22–24

Understanding

1 Is angle θ an angle of elevation or depression? Explain how you know.

2 A man standing at the top of the tower at point B, is looking at the ground at point C

a Identify the angle of depression in the figure.

b Identify the angle of elevation in the figure.

c What can be said about the size of the two angles?

8 The final approach of an aeroplane when landing requires the pilot to adjust the angle of depression to 3°, as shown in the diagram. If the plane is 12 m above the runway and has d m until touchdown, find d rounded to the nearest metre.

9 A man stands at point A looking at the top of two poles. Pole 1 has a height 8 m and an angle of elevation of 34° from point A. Pole 2 has a height 25 m and an angle of elevation of 57° from point A.

a Find the distance from A to B, rounded to two decimal places.

b Find the distance from A to C, rounded to two decimal places.

c Find BC, the distance between the two poles in metres. Round your answer to one decimal place.

10 A fighter jet, flying at an altitude of 2000 m is approaching an airport. The pilot measures the angle of depression to the airport to be 12°. One minute later, the pilot measures the angle of depression again and finds it to be 19°.

a Find the distance AC, rounded to the nearest metre.

b Find the distance BC, rounded to the nearest metre.

c Find the distance covered by the jet in that one minute.

11 A room measures 6 m in length and 5 m in width. The angle of elevation from the bottom left corner to the top right corner of the room is 60°. Rounding all answers to two decimal places:

a Find the distance from one corner of the floor to the opposite corner of the floor.

b Find the height of the room.

c Find the angle of elevation from the bottom corner of the 6 m long wall to the opposite top corner of the wall.

d Find the angle of depression from the top corner of the 5 m long wall to the opposite bottom corner of the wall.

Pole 2

Ex 2

Ex 3

16 Buzz is standing 250 m from a building and measures the angle of elevation of the top of the building to be 27°.

a If the difference in height between the top of the building and Buzz’s eye is h m, find h rounded to two decimal places.

b If Buzz’s eye is 172 cm from the ground, find the height of the building, rounded to one decimal place.

17 From the top of a rocky ledge that is 274 m high, the angle of depression to a boat is 15°. If the boat is d m from the foot of the cliff, find d rounded to two decimal places.

18 Amelia measures the angle of elevation to the top of a tree from a point, 29 m away from the base, to be 31°. Find the height of the tree, h, rounded to the nearest metre.

19 The angle of elevation from an observer to the top of a tree is 18°. If the distance between the tree and the observer is d m and the tree is known to be 3.53 m high, find the value of d rounded to two decimal places.

20 At a certain time of the day a light post, 6 m tall, has a shadow of 5.8 m. If the angle of elevation of the sun at that time is θ°, find θ rounded to two decimal places.

21 A ramp of length 311 cm needs to ascend at an angle between 10° and 20° for it to be safe to use.

a If the height of the ramp is 152 cm, find the angle the ramp makes with the ground, rounded to two decimal places.

b If the height of the ramp is 100 cm, find the angle the ramp makes with the ground, rounded to two decimal places.

c If the height of the ramp is 25 cm, find the angle the ramp makes with the ground, rounded to two decimal places.

d At which height(s) is the ramp safe?

22 Two hot air balloons are tied to the ground below, each at a different location. An observer at each location measures the angle of elevation to the opposite balloon. The observers are 1600 m apart as shown in the diagram:

a Calculate the difference in height between the two balloons. Round your answer to the nearest metre.

b Find the direct distance between the two balloons. Round your answer to the nearest metre.

23 Roald is standing at point P and observes two poles, AB and CD, of different heights. P, B, and D are on horizontal ground:

From P, the angles of elevation to the top of the poles at A and C are 29° and 18° respectively. Roald is 16 m from the base of pole AB. The height of pole CD is 7 m.

a Calculate the distance from Roald to the top of pole CD, rounded to two decimal places.

b Calculate the distance from Roald to the top of pole AB, rounded to two decimal places.

24 A ski slope is inclined at 15° to the horizontal. At the bottom of the slope, there is a starting point labeled as point A. From point A, the angle of elevation to the top of a pine tree, labeled as point D, with its base at point B on the slope, is 28°. The pine tree grows vertically from point B. After skiing 80 m along the slope and passing point B, the skier arrives at point C. From point C, the angle of elevation to point D, the top of the pine tree, is 58°.

a Find the value of ∠ADC

b Compute the distance, DA, rounded to one decimal place.

c Determine the height of the pine tree, DB, rounded to the nearest metre.

9.06 Bearings

After this lesson, you will be able to…

• use the starting direction and acute angle to describe the location of a point.

• write bearings using mathematical notation.

• apply bearings in 3D problems by drawing relevant triangles and using trigonometric ratios.

• draw a diagram of a 3D situation, find distances and calculate angles related to bearings.

Compass bearings

In surveying and navigation, bearings are used to help identify the location of an object.

The four main directions of a compass are known as cardinal directions. They are north (N), east (E), south (S), and west (W).

In between each of these cardinal direction are intermediate directions of northeast (NE), southeast (SE), southwest (SW), and northwest (NW) that are 45° from the two cardinal directions closest.

There are smaller divisions shown on the compass as well.

A compass bearing describes the location of a point using:

• The starting direction of either north or south;

• The acute angle needed to rotate

• The direction to rotate, east or west.

Starting direction Direction of rotation Angle rotated by

Apply the idea

Since A is closer to north, we start at North, rotate clockwise towards the East direction. So the letters that should be written is in option A, N ⬚ E.

b Complete the compass bearing of point A from O. N ⬚ E

Create a strategy

Write the angle measure shown in the figure.

Idea summary

Starting direction Direction of rotation Angle rotated by

Apply the idea N 35°E

A compass bearing describes the location of a point using:

• The starting direction of either north or south

• The acute angle needed to rotate

• The direction to rotate, east or west.

Three-figure (true) bearings

A three-figure bearing are:

• Measured from north (N)

• Measured in a clockwise direction

• Written using three figures

T is often, but not always, used to describe a true bearing. If the angle measure is less than 100° it would be written as 040° or 040°T.

To use true bearings to describe the location of a plane at point B from the airport at point A:

• Place the centre of a compass on the starting point, in this case the airport.

• Starting at North, rotate clockwise until we get to the line AB

• Write the angle as the true bearing of point B.

The true bearing of B from A is 127° or 127°T.

Example 3

Consider the following diagram:

a Find the true bearing of point A from point C

Create a strategy

We can find the angle between AC and north. Then, we can subtract that from 360°.

Apply the idea

To find the angle that AC makes with true north, we can extend the north line from point C to segment AB This will create a right-angled triangle.

Recall the acute angles in a right-angled triangle sum to 90°, so the angle that AC makes with true north is 90   42 = 48°.

Starting from point C and rotating clockwise from North gives us an angle of 360 48 = 312°.

This means the true bearing is 312°T.

b Find the compass bearing of point A from point C.

Create a strategy

Determine if going from C to A is north or south and east or west.

Apply the idea

From C to A it is north and west which means it is N 48°W.

Reflect and check

Notice going from A to C would be the opposite direction and would be S 48°E.

Example 4

The position of a ship S is given to be 20 kilometres from P, on a true bearing of 049° T. The position of the ship can also be given by its (x, y) coordinates.

a If the ship’s x-coordinate is x, find x rounded to one decimal place.

Create a strategy

Use the sine ratio given by: sin θ =

Apply the idea

Based from the given diagram, we are given θ = 49°, opposite = x, and hypotenuse = 20.

Substitute the values

Multiply both sides by 20

Evaluate the multiplication

Evaluate and round

b If the ship’s y-coordinate is y, find x rounded to one decimal place.

Create a strategy

Use the cosine ratio given by: cos θ =

Apply the idea

Based from the given diagram, we are given θ = 49°, adjacent = y, and hypotenuse = 20.

Substitute the values

Multiply both sides by 20

Evaluate the multiplication

Evaluate and round

Idea summary

True bearings are:

• Measured from north (N)

• Measured in a clockwise direction

Bearings in 3D

• Written using three figures

• Usually written with a T at the end

A problem in three dimensions may use bearings to describe the situation. It is helpful to first draw a diagram if one is not given to us and then solving as usual for problems in 3D by drawing out triangles in 2D.

Example 5

A miner is mining 500 m below the surface. If you walk 300 m north and then 400 m west from the office, you will be directly above the miner.

a Draw a diagram of the situation in 3D and the related triangles in 2D.

Create a strategy

We can start by drawing the surface using the bearings and then placing the miner below the surface. It is helpful to label the vertices.

Apply the idea

The whole scenario on one diagram.

We can break this into two right-angled triangles which have a common side of WO

b Find the distance from the office to the point on the surface directly above the miner.

Create a strategy

We want to find the length OW. We have the other two lengths on a right-angled triangle, so can use Pythagoras’ theorem.

Apply the idea

For triangle ONW, we know two sides and have one missing side, let’s use Pythagoras’ theorem to find OW

Pythagoras’ theorem

Substitute known values

Evaluate the indices and addition

Take the square root of both sides

Simplify

The distance from the office to the point on the surface above the miner is 500 m.

Reflect and check

We may have noticed this is the Pythagorean triple, 3, 4, 5 multiplied by 100.

c Find the direct distance from the miner to the Office, to the nearest metre.

Create a strategy

We want to find the length OM. We have the other two lengths on a right-angled triangle, so can use Pythagoras’ theorem.

Apply the idea

For triangle OWM, we know two sides and have one missing side, let’s use Pythagoras’ theorem to find OM

Pythagoras’ theorem

Substitute known values

Evaluate the indices and addition

Take the square root of both sides

Simplify the surd

Evaluate using a calculator and round

The direct distance from the office to the miner is 707 m.

Reflect and check

If we use this rounded value instead of the exact value of , there will be some error introduced.

d At what angle of depression from the office would you dig to reach the miner.

Create a strategy

We are now trying to find ∠WOM. Since this is a right-angled triangle, we can use trigonometric ratios to find this angle.

Apply the idea

We know all of the side lengths in this triangle, so we can use any pair. However, one of them is rounded, so we will use the two exact values.

The exact value sides are the opposite and adjacent.

Write the ratio

Substitute known values

Use inverse tan ratio

Evaluate using a calculator

Reflect and check

Instead of using trigonometric ratios, we could have recognised that this is an isosceles rightangled triangle, so both acute angles would be 45°.

Idea summary

Although bearings refer to directions on the surface, we may use them in 3D problems where there is point above or below ground which is involved. We can draw relevant triangles in 2D to solve for one problem at a time.

Practice questions

Choose your question path

1–6, 8, 9, 11–13, 15, 18–20, 21abc, 24, 25, 27, 29, 30

1–6, 8, 9, 11–13, 16, 18–20, 21bcd, 22, 24–27, 30

3–5, 7, 8, 10, 14, 16, 17, 20, 21cd, 22–24, 26–28, 31, 32

Understanding

1 Write the compass bearing for point P from point O a b

2 Write the true bearing of P from point O a b

3 Which of these diagrams correctly displays WNW?

4 What is the equivalent true bearing for these compass bearings?

a Southwest

c West-Northwest

e Northeast

Fluency

b Northwest

d South-Southeast

f Southeast

5 i Find the true bearing of point A from O. ii Find the compass bearing of point A from O

Ex

6 Determine the true bearing for:

a Point B from point A.

b Point C from point A

c Point C from point B.

7 Determine the true bearing for:

a Point A from point B

b Point A from point C

c Point B from point C

8 a Find the true bearing of point A from point C

b Find the compass bearing of point A from point C.

9 a Find the true bearing of point B from point C

b Find the compass bearing of point B from point C.

10 a Find the true bearing of point C from point A

b Find the compass bearing of point C from point A

11

a Find the true bearing of point C from point B.

b Find the compass bearing of point C from point B

12 If a person walks at a compass bearing of SW, and then turns around to go back in the direction they came, what direction are they heading now?

13 An aeroplane sets off on a bearing of 051°T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

14 A boat sets off on a bearing of 125°T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

15 If the bearing of the clearing from the town is a°, find a rounded to the nearest degree.

16 Luke sailed for 116 km on a bearing of 231°.

Find w, the distance in kilometres he has sailed west from his starting point, rounded to one decimal place.

17 Shortly after take-off, a plane is 42 km south and 57 km west of the airport in Sydney that it left from.

a Find the size of the angle marked b, to one decimal place.

b Find the compass bearing of the plane from the airport.

c What is the true bearing of the plane from the airport?

18 The position of a ship S is given to be 20 km from P, on a true bearing of 049° T. The position of the ship can also be given by its (x, y) coordinates.

a Find the ship’s x-coordinate rounded to one decimal place.

b Find the ship’s y-coordinate rounded to one decimal place.

19 On an orienteering course, Beth runs 1000 metres north from point A to point B, then turns east and runs to point C. If the true bearing of C from A is 062°T and the direct distance Beth must run to get back to point A is d metres, find d to the nearest metre.

20 From a point 15 m due north of a tower, the angle of elevation of the tower is 32°.

a Find the height of the tower h. Round your answer to two decimal places.

b Find θ, the angle of elevation of the tower at a point 20 m due east of the tower. Round your answer to the nearest degree.

Reasoning

21 Given that a, b, c and d are angles between 0° and 90°:

a If a compass bearing is N a°W, write an expression for the true bearing.

b If a compass bearing is S b°W, write an expression for the true bearing.

c If a compass bearing is S c°E, write an expression for the true bearing.

d If a compass bearing is S (90  a) °E, write an expression for the true bearing.

22 a Explain why the overall horizontal distance from A to C is equal to 12 cos 45° + 16 cos 30°.

b Write an expression using sin which will calculate the overall vertical distance from A to C

c Use Pythagoras’s theorem to calculate the direct distance from A to C

d Find the bearing of A from C.

23

a Find the direct distance from A to C.

b Find the bearing of A from C

Problem-solving

24 During a group hike, a hiker walked 11 km NE of base camp. In the evening the hiker was separated from the group and after a night lost in the wilderness, the hiker was found 11 km NW of the base camp. What compass bearing must the hiker have followed during the night?

25 During a rescue search, a helicopter flew west from point X to point Y, then changed course and flew 10.7 km north to point Z.

If point Z is on a bearing of 335° T from point X:

a Find the size of ∠Y XZ

b Find b, the distance from point Y to point X, rounded to one decimal place.

c Calculate d, the distance that the helicopter must fly between point Z and point X, rounded to one decimal place.

26 In remote locations, photographers must keep track of their position from their base. One morning a photographer sets out from base, B, to the edge of an ice shelf, S, on a bearing of 055°. She then walked on a bearing of 145° to point P, which is 916 m due east of base.

Rounding all answers to one decimal place:

a State the size of ∠BSP.

b Find the distance BS

c Find the distance SP.

d If the photographer walks directly back to her base from point P, determine the total distance travelled.

27 A fishing boat in search of large schools of fish sails from point A for 26 km on a bearing of 205° to point C. It then sails a further 24 km on a bearing of 226° to point E. Rounding all answers to one decimal place:

a Find the size of the angles: i ∠BAC ii ∠DCE

b Find the distance the boat is from its starting point as a straight line from A to E

c Find θ°, the true bearing of the boat from its starting point, to the nearest degree.

28 A drone travels due east for 2.2 km and then travels on a bearing of S 32°E for 5.9 km.

a Given that the compass bearing is a°, write the bearing needed to return to the starting point, in terms of a.

b Find x, the distance between the end point to the start point of the drone’s flight. Round your answer to two decimal places.

c Find a rounded to the nearest degree.

29 A yacht sailed in a direction so that its final position was 248 km west and 225 km south of its starting point.

a Draw a diagram to represent the situation.

b Find the true bearing on which the yacht sailed, rounded to one decimal place.

c Find the total distance the boat has sailed, rounded to one decimal place.

30 Three television presenters are practising their navigation skills before heading off on an expedition to a remote location.

Belinda at point B is positioned 17.6 m south of Amelia at point A. Carl at point C is due east of Belinda and on a bearing of S 38°E from Amelia.

a Draw a diagram to represent the situation.

b If Amelia and Carl are d m apart, find d to one decimal place.

31 A yacht travelled a distance of 30 km from port A at a bearing of 110°, then another 40 km to port C at a bearing of 130°. What is the distance between ports A and C?

32 A and B are two positions on level ground. From an advertising balloon at a vertical height of 740 m, A is observed at a bearing of 043° and B at a bearing of 133°. The angles of depression of A and B, as viewed from the balloon, are 30° and 20° respectively.

Rounding all answers to the nearest metre:

a Find the distance from point A to the point directly below the advertising balloon.

b Find the distance from point B to the point directly below the advertising balloon.

c Find the distance between points A and B.

9.07 Optional: Sine and cosine

rules

After this lesson, you will be able to…

• utilise the sine rule to find unknown sides and angles.

• distinguish between the two forms of the sine rule and when to use each.

• identify the conditions that lead to multiple solutions when using the sine rule.

• define and explain the cosine rule and its applications in solving problems related to non-right-angled triangles.

• use the cosine rule to find unknown sides or angles.

Sine Rule

The sine rule is used in non-right-angled triangles that involve two sides and two angles. We are able to use this rule if we want to find an unknown side or angle.

Interactive exploration

Explore online to answer the questions

mathspace.co

Use the interactive exploration in 9.07 to answer these questions.

1. Which side (shortest, middle, longest) is across from the largest angle? Which side is across from the smallest angle?

2. What do you notice when you change the value of the angles (using angle C)? What changes? What stay the same?

3. What happens to the value of sin (C) when it is C = 90°?

4. What happens when C is 90° or 180°?

The sine rule, sometimes called the law of sines or sine law, tells us that:

A, B, C the measures of the angles in the triangle a, b, c the sides in the triangle

We can also take the reciprocal of each fraction to give the alternate form:

This alternate form is used to find angles. Suppose we know the side lengths a and b and the angle B, and we want to find the angle A.

Using the form of the sine rule sin , we can solve for A

We can make sin A the subject by multiplying both sides by a:

Then, to find B itself, we need to take the inverse on both sides:

Example 1

Find the side length a using the sine rule. Round your answer to two decimal places.

Create a strategy

Use the sine rule: .

Apply the idea

Substitute b = 18, A = 33, B = 69

Multiply both sides by sin 33°

Evaluate

Example 2

Find the value of the acute angle A using the sine rule. Round your answer to two decimal places.

Create a strategy

Use the sine rule:

Apply the idea

Substitute a = 98, b = 81, B = 50

Multiply both sides by 98

Evaluate

Take the inverse sine of both sides

Evaluate

Idea summary

The sine rule can be written in two forms. The first form is best when solving for an angle. The second form is best when solving for a side.

A is an angle of a triangle opposite side a B is an angle of a triangle opposite side b C is an angle of a triangle opposite side c

We will generally only use two angle-side pairs when solving. We need one side-angle pair both known and one other piece of information.

Ambiguous case of the sine rule

When we are solving for angles using the sine rule, there may be two possible solutions.

Interactive exploration

Explore online to answer the questions

Use the interactive exploration in 9.07 to explore the number of possible solutions when the information given.

The ambiguous case occurs when it is possible to have multiple solutions with the information given. When using the sine rule, it is possible to have no solution, one solution or two solutions:

• Case 1: No triangle exists

This is the case when an error occurs in the calculator while solving for the unknown angle. This means that no such triangle with the given side lengths and angle exists.

• Case 2: Exactly one triangle exists

This is the case when the sum of the given angle and the supplement to the solution calculated is equal to or exceeds 180°.

• Case 3: Two possible triangles exist

This is the case when the sum of the given angle and the supplement to the solution are less than 180°. Then, the possible triangles are as follows:

• A triangle that includes the calculated solution

• A triangle that includes the supplement to the calculated solution

To find the second possible solution, subtract the first value of B that you found from 180°.

Example 3

Find the value of x in degrees, given that x is obtuse. Round your answer to two decimal places.

Create a strategy

Use the sine rule: .

Apply the idea

Multiply both sides by 18.8

Evaluate

Take the inverse sine of both sides

Evaluate and round

To find the obtuse angle x we subtract the acute angle from 180°:

x = 180   39.09

Subtract 39.09 from 180 = 140.91° Evaluate

Example 4

If possible, solve the triangle where a = 85, b = 93, and ∠A = 61°. Round your answers to two decimal places.

Create a strategy

Draw and label a diagram of the triangle; it does not need to be drawn to scale.

From this diagram we can see that we were given two sides and their non-included angle. This is an example of the ambiguous case and when we solve we may find one, two, or no solutions to this problem.

We are given a pair of opposite sides and angles with side a and angle A in addition to side b. We will set up the equation:

to find ∠B.

Apply the idea

Sine rule

Substitute given values

Multiply both sides by 93

Apply the inverse sine to both sides

Evaluate and round

Now that we’ve found an acute solution to the problem, we need to find its supplement and check if this will produce a valid second solution.

The supplement of ∠B1 = 73.12° is ∠B2 = 180°   73.12° = 106.88°. If we add these two angles together we get 61° + 106.88° = 167.88° which does not exceed the maximum 180° possible in a triangle. That means this triangle has two solutions.

Solution 1: ∠B1 = 73.12°

So far we have a = 85, b = 93, ∠A = 61° and ∠B1 = 73.12°. Use the triangle angle sum to solve for the missing angle, ∠C1 = 180°   (61° + 73.12°) = 45.88° and use the sine rule to find the missing side, c1:

Sine rule

Substitute known values

Multiply both sides by sin 45.88°

Evaluate and round

The three side lengths are a = 85, b = 93, and c1 = 69.77 and the three angles are ∠A = 61°, ∠B1 = 73.12° and ∠C1 = 45.88°.

Solution 2: ∠B2 = 106.88°

For the second solution we have a = 85, b = 93, ∠A = 61° and ∠B2 = 106.88°. Use the triangle angle sum to solve for the missing angle, ∠C2 = 180°   (61° + 106.88°) = 12.12° and use the sine rule to find the missing side, c2:

Sine rule

Substitute known values

Multiply both sides by sin 12.12°

Evaluate and round

The three side lengths are a = 85, b = 93, and c2 = 20.40 and the three angles are ∠A = 61°, ∠B2 = 106.88° and ∠C2 = 12.12°. The diagram shows a visual of both solutions:

Reflect and check

There will be no solution, or a calculator error, if we attempt to take the inverse sine of a value greater than 1. There will be exactly one solution if the solved angle is acute, but its supplement is too large to create a second valid triangle.

9.07 Optional: Sine and cosine rules

Idea summary

Sine rule (ambiguous case)

If you are trying to find an angle using the sine rule (with two known sides and a known angle), use the following rule to find one value of B:

If the side opposite the known angle is the shorter side, you are in the ambiguous case. Subtract the first value of B that you found from 180° to find another possible solution. Check that the second solution is a valid solution using the sum of angles in at triangle.

Cosine Rule

The cosine rule rule is used in non-right-angled triangles that involve three sides and one angle. We are able to use this rule if we want to find an unknown side or angle.

Two side lengths and one angle, opposite the unknown side, are known.

Three side lengths are known, and one angle is unknown.

The cosine rule says:

To find a missing angle using the cosine rule, we need to know all three side lengths. When looking to solve for an unknown angle, the equation can also be rearranged and written as

Example 5

Find the length of a using the cosine rule. Round your answer to two decimal places.

Create a strategy

Use the cosine rule a 2 = b2 + c 2  2bc cos A

Apply the idea

Write the formula

Substitute b = 13, c = 30, A = 33

Evaluate

Take the square root of both sides and round

Example 6

Find the value of angle w using the cosine rule. Round your answer to two decimal places.

Create a strategy

Use the rearranged formula of the cosine rule given by , where w corresponds to C

Apply the idea

Write the formula

Susbtitute the values

the inverse cosine of both sides

and round

Idea summary

If we have a problem involving two sides and the included angle of a non-right angled triangle, we can use the cosine rule to find the unknown side length:

c 2 = a 2 + b2  2ab cos C

c is the side opposite angle C

a is another side length

b is another side length

If we have a problem involving 3 side lengths of a non-right angled triangle, we can use the cosine rule to find the unknown angle:

c is the side opposite angle C

a is another side length

b is another side length

Practice questions

Choose your question path

1, 3, 4, 5acdg, 6, 7, 9, 10ab, 11abdg, 13a, 14ab, 15, 16ace, 18, 20abce, 21, 23, 24, 29, 30, 32, 39, 40

1, 3, 4, 5aceg, 7–9, 10ab, 11bdfj, 13b, 14cd, 15, 16bdf, 19, 20cdef, 21, 23–25, 30–33, 37–39

2–4, 5bdf, 8, 9, 10d, 11ij, 12, 14d, 16ef, 17, 19, 20dfg, 22, 25–28, 33–37, 41–43

Understanding

1 Which strategy would you use to find the unknown: Pythagoras’ theorem, trigonometric ratios, sine rule or cosine rule?

2 For each triangle, determine if there is enough information to find all the remaining sides and angles in the triangle using the: i Sine rule ii Cosine rule

a Three sides are known.

b Two of the angles and the side included between them are known.

c Two of the angles and a side not d Two of the sides and an angle included included between them are known. between them are known.

3 Write an equation relating the sides and angles using the sine rule.

4 Write an equation relating the sides and angles using the cosine rule.

5 Find the unknown side, rounded to two decimal places. a b c d e f g h

6 Find the length, rounded to two decimal places. a HK b KJ

7 For triangle ABC, where BC = 37.8 m, ∠BAC = 57° and ∠ACB = 42°, calculate the length of side AB, rounded to two decimal places.

8 For triangle PQR, where QR = 4.79 mm, ∠QPR = 33° and ∠PRQ = 124°, calculate the length of side PQ, rounded to two decimal places.

9 For triangle ABC with ∠C = 72.53° and ∠B = 31.69°, and one side length a = 5.816 m, find:

a ∠A, rounded to two decimal places.

b The length of side b, rounded to three decimal places.

c The length of side c, rounded to three decimal places.

10 Solve for x, given that all equations relate to acute-angled triangles, rounded to two decimal places.

a b c d

11 For each triangle, find the value of the variable using the sine rule. Round your answers to one decimal place.

9.07 Optional: Sine and cosine rules

Ex

12 For triangle MNP, where NP = 100 m, MN = 204 m and ∠NPM = 58°, calculate the size of the acute angle ∠NMP, rounded to the nearest degree.

13 For each triangle, calculate ∠B rounded to the nearest degree.

a △ABC where ∠A = 57°, side a = 156 cm and side b = 179 cm

b △ABC where ∠A = 48°, side a = 2.7 cm and side b = 1.9 cm

14 Solve for θ, given that all equations relate to acute-angled triangles, rounded to two decimal places.

a b c d

15 Which angle would need to be given to find the length of AC using the cosine rule?

16 For each triangle, find the value of the variable using the cosine rule. Round your answers to two decimal places.

17 In △ABC, , find the exact length of side AB in centimetres.

18 Triangle QUV has side lengths v = 7, u = 10 and an angle at Q such that cos . Find the value of q, where q is the length of the side opposite angle Q

19 Triangle QUV has side lengths q = 63, u = 36 and ∠V = 116°. Find the value of v, where v is the length of the side opposite ∠V. Round your answer to two decimal places.

20 For each triangle, find the value of the variable in degrees. Round your answers to two decimal places.

21 Triangle QUV has side lengths q = 5, u = 10 and v = 12. Find the size of ∠V, where V is the angle opposite the side of length v. Round your answer to the nearest degree.

22 Triangle QUV has side lengths v = 23.5, u = 15.6 and q = 10.0. Find the size of ∠Q, where Q is the angle opposite the side of length q. Round your answer to the nearest degree.

Reasoning

23 Blain and Odett are trying to find side c in △ABC where a = 15 m, b = 12 m and ∠C = 34°. Review the work, explain the errors (if any) and determine the correct value for the side c Blain’s work: Odett’s work: 1 2 3

24 A rhombus of side length 10 cm has a longer diagonal of length 16 cm.

a Describe two ways you could solve for d, the length of the shorter diagonal of the rhombus.

b Solve for d

25 A teacher is writing exam questions for her maths class. She draws a triangle, labels the vertices A, B and C and labels the opposite sides a = 5, b = 8 and c = 15 respectively. She wants to ask students to find the size of ∠A. Explain what the problem is with her question.

26 Use the cosine rule to prove that any angle in a triangle with three equal sides is equal to 60°.

27 After two meteoroids collide at point A, one starts travelling in the direction of point B, while the other starts travelling in the direction of point C, with an angle of 53° between the two directions. The meteoroid projected in the direction of B is moving at a speed of 7860 km/h, while the other is moving at a speed of 10 170 km/h.

a Describe a plan to find the distance between the two meteoroids 29 minutes after the collision.

b Find the distance between the two meteoroids 29 minutes after the collision, rounded to two decimal places.

c Write an equation to find the distance between the two meteoroids m minutes after the collision. Let d be the unknown distance.

28 A pole is seen by two people, Jenny and Matt:

a Matt is x m from the foot of the pole. Find x rounded to the nearest metre.

b Find the height of the pole h rounded to the nearest metre.

Problem-solving

29 Find the value of θ, rounded to two decimal places.

30 A radio signal is sent from a transmitter at tower T, via a satellite S, to a town W, as shown in the diagram. The town is 526 kilometres from the transmitter tower. The signal is sent out from the transmitter tower at an angle of 18°, and the town receives the signal at an angle of 26°.

a Find ∠WST

b Find the distance, SW, that the signal travels from the satellite to the town. Round your answer to the nearest kilometres.

c If the satellite is h kilometres above the ground, find h. Round your answer to two decimal places.

31 Two hikers on a large flat national park walk 10 km from point D on a bearing of S 40°E to a point E. They then walk on a bearing of N 30°E to point F. If the distance DF is 15 km, find the bearing of D from F.

32 Two wires help support a tall pole. One wire forms an angle of 36° with the ground and the other wire forms an angle of 70° with the ground. The wires are 29 m apart.

a Find a, the angle made between the two wires at the top of the pole.

b Find d, the length of the longest wire in metres. Round your answer to two decimal places.

c Find h, the height of the pole. Round your answer to two decimal places.

33 Mae observes a tower at an angle of elevation of 12°. The tower is perpendicular to the ground. Walking 67 m towards the tower, she finds that the angle of elevation increases to 35°.

a Find ∠ADB.

b Find the length of the side a. Round your answer to two decimal places.

c Find the height, h, of the tower. Round your answer to one decimal place.

34 Prove the sine rule using these steps:

a For △ACD, find an expression for sin A

b For △BDC, find an expression for sin B.

c Solve for x in the equation from part (b).

d Substitute your expression for x into the equation from part (a) to prove the law of sines.

35 How many triangles are possible given this information? Explain your reasoning. Assume a is the side length opposite ∠A, b is the side length opposite ∠B.

a a = 29, b = 28, and ∠A = 37°

b a = 38, b = 29, and ∠B = 45°

c a = 54, b = 65, and ∠A = 60°

36 The sides of a triangle are in the ratio 4: 7: 8. Find the size of x, the smallest angle in the triangle. Round your answer to the nearest degree.

37 For triangle △QUV where the side lengths q, u and v appear opposite the angles Q, U and V. If q = 16, sin V = 0.5 and sin Q = 0.8. Solve for v

38 Two trains leave a station at the same time. Train H travels at an average speed of 80 km/h on a bearing of 035°T, while Train I travels at an average speed of 90 km/h on a bearing of 120°T. Find the distance between the two trains after 4 hours. Round your answer to the nearest kilometre.

39 During football training, the coach marks out the perimeter of a triangular course that players need to run around. The diagram shows some measurements taken of the course, where side length a = 14 m:

a Find ∠A.

b Find the length of side b, rounding your answer to two decimal places.

c Find the length of side c, rounding your answer to two decimal places.

d Each player must sprint one lap and then jog one lap around the triangle. This process is to be repeated 3 times by each player. If Tara can run 280 m/min, and can jog at half the speed she runs, calculate the time this exercise will take her, rounding your answer to one decimal place.

40 In a game of pool, a player has one last ball to sink into a corner pocket. The player must use a cue (stick) to hit the white ball so that it knocks the purple ball into the corner pocket.

The player judges that the white ball is about 1.5 cm away from the corner pocket, and that the distance between the two balls is about 0.9 cm, while the purple ball is also 0.9 cm from the corner pocket.

The player wants to find the angle θ which is needed to knock the white ball against the purple ball. Round your answer to two decimal places.

41 Two straight paths to the top of a cliff are inclined at angles of 23° and 20° to the horizontal:

a If path A is 108 m long, find the height h of the cliff, rounded to the nearest metre.

b Find the length x of path B, rounded to the nearest metre.

c Let the paths meet at 49° at the base of the cliff. Find their distance apart, y, at the top of the cliff, rounded to the nearest metre.

42 Two buoys, A and B, are observed from a lookout at the top of a 130 m high cliff. The bearing of buoy A is 337° and its angle of depression is 3°. The bearing of buoy B is 308° and its angle of depression is 5°:

a Find the distance, rounded the nearest metre, from buoy A to the base of the cliff.

b Find the distance from buoy B to the base of the cliff, rounded to the nearest metre.

c Find the distance between the two buoys, rounded to the nearest metre.

43 From a point A, due north of a tower, the angle of elevation to the top of the tower is 51°. From a point B, which is on a bearing of 120° from the tower and is 100 m away from point A, the angle of elevation is 25°.

Find the height of the tower rounded to the nearest metre.

Chapter 9 review

1 Find the length of the unknown side as an exact value: a

b A right-angled triangle has a hypotenuse of length 26 and another side of length 24.

c A right-angled triangle has a one side of length 6 and another side of length 5.

2 Could a triangle with side lengths of 6 mm, 8 mm, and 13 mm be a right-angled triangle? Explain.

3 Rachel hikes south of her starting point for 834 m, then turns and walks 691 m east, before stopping for a lunch break. She then travels south again for 427 m before arriving at her final destination.

a What is the shortest distance between where Rachel started and where she stopped for lunch? Round your answer to two decimal places.

b What is the shortest distance between where Rachel started and where she finished her journey? Round your answer to two decimal places.

4 A rectangular prism has dimensions as labelled on the diagram. Find each length in exact form:

5 A triangular prism has dimensions as shown. Explain the process for finding:

11 Find the value of the variable(s), rounded to the nearest whole number: a b

12 Consider the rectangular prism:

a Find the length x.

b Find the length of the prism’s diagonal y

c Find angle θ to the nearest degree.

13 A, C, and X are three points in a horizontal plane and B is a point vertically above X as shown: Find these lengths to the nearest metre: a BC b XB

14 A pyramid has a square base with side length 8 m and a vertical height of 25 m.

a Find the length of the edge AE using two different methods.

b If θ is the measure of the surface ∠BEA, find θ rounded to the nearest degree.

15 The person in the picture sights a pigeon above him. Find the angle of elevation, θ, to two decimal places.

16 An aeroplane is flying at an altitude of 3000 m and approaching an airport. The pilot measures the angle of depression to the airport to be 14°. One minute later, the pilot measures the angle of depression again and finds it to be 16°.

a Find the distance AC, to the nearest metre.

b Find the distance BC, to the nearest metre.

c Find the distance covered by the jet in that one minute, to the nearest metre.

17 At a certain time of the day a light post, 5 m tall, has a shadow of 9.4 m. If the angle of elevation of the sun at that time is θ°, find θ rounded to two decimal places.

18 Dave is standing 270 m from a building and measures the angle of elevation of the top of the building to be 19°.

a If the difference in height between the top of the building and Buzz’s eye is h m, find h rounded to two decimal places.

b If Dave’s eye is 161 cm from the ground, find the height of the building rounded to one decimal place.

19 i Find the true bearing of point A from O

ii Find the compass bearing of point A from O

20 a Find the true bearing of point A from point C

b Find the compass bearing of point A from point C

21 If a person walks at a compass bearing of NE, and then turns around to go back in the direction they came, what direction are they heading now?

22 A boat sets off on a bearing of 121 °T. After some time, it needs to turn back and head to its original position. On what true bearing must it travel?

23 Uri sailed for 155 km on a bearing of 215°. Find w, the distance in kilometres he has sailed west from his starting point, rounded to one decimal place.

24 During a group hike, a hiker walked 12 km NE of base camp. In the evening the hiker was separated from the group and after a night lost in the wilderness, the hiker was found 12 km NW of the base camp. What compass bearing must the hiker have followed during the night?

25 (Optional) Find the unknown, rounded to two decimal places.

26 (Optional) For each triangle, calculate ∠B to the nearest degree.

a △ABC where ∠A = 48° side a = 2.7 cm and side b = 1.9 cm

b △ABC where ∠A = 57° side a = 156 cm and side b = 179 cm

27 (Optional) To use the cosine rule to find the length of AC, which angle would need to be given?

28 (Optional) Find a using the cosine rule. Round your answers to two decimal places.

29 (Optional) Triangle ABC has side lengths a = 83, b = 45 and c = 110. Find ∠C. Round your answer to the nearest degree.