KT Classroom: 3 Dimensional Trigonometry

MATHEMATICS Learner’s Study and Revision Guide for Grade 12 3‐D TRIGONOMETRY

Revision Notes, Exercises and Solution Hints by

Roseinnes Phahle

Examination Questions by the Department of Basic Education

Preparation for the Mathematics examination brought to you by Kagiso Trust

Contents Unit 18 Revision notes

3

Exercise 18.1

4

Answers

5

6

Examination questions with solution hints and answers

11

More questions from past examination papers

15

Answers

21

Strategy for solving 3‐dimensional problems

How to use this revision and study guide 1. Study the revision notes given at the beginning. The notes are interactive in that in some parts you are required to make a response based on your prior learning of the topic in class or from a textbook.

2. “Warm‐up” exercises follow the notes. Some exercises carry solution HINTS in the answer section. Do not read the answer or hints until you have tried to work out a question and are having difficulty.

3. The notes and exercises are followed by questions from past examination papers.

4. The examination questions are followed by blank spaces or boxes inside a table. Do the working out of the question inside these spaces or boxes.

5. Alongside the blank boxes are HINTS in case you have difficulty solving a part of the question. Do not read the hints until you have tried to work out the question and are having difficulty.

6. What follows next are more questions taken from past examination papers.

7. Answers to the extra past examination questions appear at the end. Some answers carry HINTS and notes to enrich your knowledge.

8. Finally, don’t be a loner. Work through this guide in a team with your classmates.

Three dimensional trigonometry

REVISION UNIT 18: THREE DIMENSIONAL TRIGONOMETRY Dimension • • •

A line is ONE dimensional.

A plane or surface like the paper you are writing on is TWO dimensional.

Space like your classroom taken as a whole to include the floor, walls and roof is THREE dimensional.

Another example which is a common one in 3‐D problems is a pole or mast vertically erected on horizontal ground and supported by ropes or wires (called stays) stretching at a slant from the top of the pole to the ground – see if you can illustrate this on paper!

Angles of elevation and depression

In the space opposite illustrate and label an angle of elevation and an angle of depression

Rules and formulae Write down and illustrate with diagrams in the spaces below all the formulae and rules you have learnt in geometry and trigonometry that relate to triangles:

Rule/formula

Illustration with appropriate triangle

1.

Pythagoras Theorem

2.

Basic trigonometric formulae:

• • •

Sin θ = Cos θ = Tan θ =

3.

Sine rule:

4.

Cosine rule:

5.

Area of a triangle in terms of sine:

NOTE: all the above rules and formulae apply to a triangle and are all you need to know to solve three dimensional problems. So in solving 3‐D problems we must apply them to triangles that form the surfaces of a 3‐D figure. This means you must extract triangular planes from the 3‐D figure and apply the rules to the triangles.

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Exercise 18.1 Solve each of the following triangles (this means evaluate the lengths and angles which are not given) and also evaluate the area in each case: 1.

3.

2.

4.

5.

6.

7.

8.

Three dimensional trigonometry

ANSWERS EXERCISE 18.1

1. BC = 36km

ˆ = 38 5. A o

ˆ = 28 B o

ˆ = 60 B o

ˆ = 56 C Area = 252,5 sq km

2. BC = 46km

ˆ = 82 C Area = 156 sq m

ˆ = 36o B ˆ = 39o C

B = 45

o

o

ˆ = 72 6. A o

o ˆ ˆ = 63o C

Area = 85,6 sq cm

Area = 405,7 sq m

ˆ = 26 3. B o

7. A = 47

ˆ

ˆ = 65 A BC = 29m Area = 203 sq m

o

C = 78 AB = 11cm Area = 35,2 sq cm

8. BC = 40cm

ˆ

ˆ = 41o 4. B

ˆ = 39 C AC = 24cm Area = 271,6 sq cm

o

o

5

o

ˆ = 11 B ˆ = 63o C Area = 141 sq cm

o

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STRATEGY FOR SOLVING 3‐DIMENSIONAL PROBLEMS To solve three‐dimensional problems, it is important to be able to visualize TRIANGLES contained in the diagram. Then redraw the TRIANGLES in two‐dimensions marking in the angles and lengths of sides given and use an appropriate TRIGONOMETRIC RATIO, COSINE RULE, SINE RULE or AREA FORMULA to obtain the answer.

Example:

In the figure shown opposite, OB represents the height of a pyramid that is 10 metres high. The angle of elevation of the top of the pyramid is 35 o from the point S, and 50 o from the point K. Find:

a) the length of KS;

b) the area of ΔBKS ;

ˆ S . c) using the sine rule, the angle BK

Solution:

Our solution is rather long winded because it is trying to engage you in a discussion of how to go about seeking a solution to 3‐dimensional problems. Bear in mind the strategy outlined above of extracting triangles from the 3‐D figure. a) Finding the length of KS KS appears in two triangles: Δ s BKS and OKS. We thus need to solve one or the other to evaluate the length of KS. Which of the triangles will it be?

These are not right‐angled triangles. So we cannot apply Pythagoras’ Theorem. Nor can we use the definitions of sine, cosine and tangent in these two triangles.

To find out what to do, extract and in the boxes below display the triangles making up the 3‐D figure:

ΔBOK

ΔBOS

ΔBKS

ΔOKS

Three dimensional trigonometry

Let’s see if we can use Δ BKS to work out the length of KS. We can find the lengths of BK and BS using Δ s BOK and BOS. Then what?

However, this does not put us in a position to find the length of KS. To find KS we will need to know the

ˆ in order to apply the cosine rule, or know the sizes of two angles in order to apply the sine size of B ˆ and another angle? None. rule. But what information in Δ BKS do we have to calculate B So we cannot use Δ BKS. The only option we have to calculate the length of KS is Δ OKS. From Δ BOK:

tan 50 o =

OK =

OB 10

= OK OK

10 = 8,39 tan50 o

From Δ BOS:

tan 35 o =

OS =

OB 10

= OS OS

10 = 14,28 tan35 o

Now we are in a position to apply the cosine rule:

ˆS KS2 = OK 2 + OS2 − 2.OK.OS.cos KO

KS 2 = 8,39 2 + 14,28 2 − 2 × 8,38 × 14,28 × cos 60 o

KS 2 = 154,5013

KS = 12,43m

b) Finding the area of Δ BKS The area of Δ BKS will be given by any of the formulae:

1 ˆ or 1 KB.BS.sinB ˆ or 1 BS.SK.sinSˆ BK.KS.sin K 2 2 2

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Apart from KS, we do not know the measurements of the other sides nor do we know the size of any angle. Using Δ s BOK and BOS we can find the measurements of the sides KB and SB:

sin 50 o =

10 10 so that KB = = 13,05 KB sin50 o

sin 35 o =

10 10 so that SB = = 17,43 SB sin35 o

And

By now we know all the sides of Δ BKS. All we need to find in order to apply any of the formulae above is the size of any one of the angles. For this purpose, we use the cosine rule:

ˆ S KS2 = KB2 + SB2 − 2.KB.BS.cosKB

ˆ S 12,43 2 = 13,05 2 + 17,43 2 − 2 × 13,05 × 17,43 × cos KB

ˆ S 154,5049=170,3025+303,8049‐454,9230 cos KB

ˆ S 154,5049=474,1074‐454,9230 cos KB

ˆ S = 474,1074‐154,5049 = 319,6025 454,9230 cos KB

ˆ S = 319,6025 = 0,7025 cos KB 454,9230

ˆ S = cos -1 (0,7025 ) = 45,37 o KB

ˆ S in much fewer steps than we have done above. But we chose to Note: We could have worked out KB do it this way in order to show you that if you have difficulty in transposing a formula then substitute into it and proceed in a step by step fashion to get what you want.

Note also: We carried more decimal places than we needed in the answer expressed to only two decimal places. The rule is that you must not round off until you have arrived at the answer. But, of course, this entire calculation could be carried out on your electronic calculator without puting pen to paper. Try doing this and see if you get the same answer.

Now we can find:

area of Δ BKS =

1 ˆ = 1 × 13,05 × 17,43 × sin 45,37 o = 80,94 sq m KB.BS.sinB 2 2

Three dimensional trigonometry

ˆS c) Using the sine rule to find BK

ˆ S sin KB ˆS sin BK

= BS KS

ˆ ˆ S = BS × sinKBS = 17,43 × sin 45,37 = 0,998 sin BK KS 12,43

ˆ S = sin -1 (0,998 ) = 86,38 o BK

Example: (This example is taken from the Free High School Science Texts, Grade 12 Mathematics, which is a textbook freely available on the internet at www.fhsst.org and includes textbooks for Grades 10 and 11 as well).

D is the top of a tower of height h . Its base is at C .

The triangle ABC lies on the ground (a horizontal

plane). If we have that BC = b , DBˆ A = α ,

DBˆ C = x and DCˆ B = θ , show that

h =

b sin α sin θ

sin (x + θ )

Solution: Extract triangles from the 3‐D figure. These are Δ s ABC, ADC, ADB and DBC. Show these triangles in the boxes below marking in the sides, right angles and the other angles indicated in the diagram above.

Δ ABC

Δ ADC

Δ ADB

Note that AD ⊥ ar AB because AD ⊥ ar to the horizontal plane ABC.

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Δ DBC

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Note also that the expression we are required to prove does not contain squares or the tangent function thus ruling out the use of the cosine rule or tan.

Looking at Δ ABC: b is the only measurement this triangle contains so it is of no use as a start.

Looking at Δ ADC: One side and a right angle is known in this triangle. But without knowing the sizes of the other angles, we cannot apply the definitions of sine, cosine or tangent. So this triangle is also of no use in proving the expression we are asked to prove.

h

BD

Looking at Δ ADB:

sin α =

Looking at Δ DBC:

ˆ C = 180 - (x + θ ) BD

or

{

}

h = BDsinα

ˆ C = sin 180 o − ( x + θ ) = sin ( x + θ ) hence sin BD

Why?

Applying the sine rule:

BD b DC = =

ˆ C sinx sin θ sin BD

(we can discard the last bit because what we need is BD to be able to substitute in h = BDsinα ).

BD b =

sinθ sin (x + θ )

BD =

so that

h = BDsinα =

h=

b sin θ

sin ( x + θ )

b sin α sin θ

sin ( x + θ )

b sin θ . sin α sin ( x + θ )

Three dimensional trigonometry

PAPER 2 QUESTION 7

DoE/ADDITIONAL EXEMPLAR 2008

11

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PAPER 2 QUESTION 7

Number Hints and answers 7.1 There is no right angle in this triangle and therefore no hypotenuse. So you cannot use the simple sine and cosine ratios that depend on the hypotenuse.

Rather you must use either the sine or cosine rule or both.

You decide which to use.

Remember you are not to use a calculator.

7.2.1 Extract the right angled‐ triangles in which the sides AC and AD are found.

It will help to draw these triangles labeling the vertices as in the 3‐D figure given and indicating the right angles and known lengths to the sides.

Answers: AC = 61m; AD = 67,3m 7.2.2 Extract the relevant triangle that has all the information you need to calculate the length CD.

Decide on the formula to use.

Answer: CD = 74,68m 7.2.3 Decide on the formula to use to calculate the area.

Answer: Area of Δ ACD = 1940,82sq m

DoE/ADDITIONAL EXEMPLAR 2008

Work out the solutions in the boxes below

Three dimensional trigonometry

PAPER 2 QUESTION 7

DoE/NOVEMBER 2008

13

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PAPER 2 QUESTION 7

Number Hints and answers

7.1 3‐D problems in trigonometry reduce to solving for the angles and sides of triangles. If there is a right angle in the triangle then use the definitions of sin, cos or tan whichever is appropriate to get a solution. If there is no right angle, use either the sine or cosine rule whichever is appropriate.

Wihich triangle are you going to use to calculate the length of LB?

Answer: LB = 3,58 m

7.2 There is no right angle in this triangle and therefore no hypotenuse. So you cannot use the simple sine and cosine ratios that depend on the hypotenuse.

Rather you must use either the sine or cosine rule or both.

You decide which to use.

Answer: AB = 7,3 m or 7,4 m

7.3 Decide on the formula to use to calculate the area.

Answer: 8,5 m or 8.6 m

DoE/NOVEMBER 2008

Work out the solutions in the boxes below

Three dimensional trigonometry

MORE QUESTIONS FROM PAST EXAMINATION PAPERS Exemplar 2008

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Preparatory Examination 2008

Three dimensional trigonometry

Feb – March 2009 Question 7

17

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November 2009 (Unused paper)

Three dimensional trigonometry

November 2009 (1)

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Feb – March 2010

Three dimensional trigonometry

ANSWERS Exemplar 2008 7.1 BD = 28,6m 7.2 Proof required. Provide a proof and check with the teacher if it is correct.

7.3 Area of Δ BEC = 160,4 sq m

Preparatory Examination 2008 8.1 OC =

h

tan y

8.2 Proof required. Provide a proof and check with the teacher if it is correct.

8.3 Area = 1,93 sq units

Feb/March 2009 7.1 PB = 22,65m 7.2 PA = 19,55m 7.3 AB = 15,40m

November 2009 (Unused paper) 7.1.1 QR =

)

7.1.2 Proof required. Provide a proof and check with the teacher if it is correct.

7.1.3 α = 58,56 o

7.2 Area = 1688,75 sq cm

November 2009(1)

ˆ A = 58,58 o 11.1.1 MB 11.1.2 EA = 102,41km

11.2 Proof required. Provide a proof and check with the teacher if it is correct.

Feb/March 2010 12.1 Proof required. Provide a proof and check with the teacher if it is correct.

12.2 DE = 1559,50m

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(

6 cosα + 3sinα

sinα