Let Us Recall! Phytagoras Theorem: c² = a² + b² Trigonometry:
c
a
a sin θ = c cos θ =
θ b
b c
a tan θ = b
How To Measure The Distance? House A
Distance?
House B
How about this? House A House C Distance, m
Distance, n Distance?
House B
How about this? House A House C Distance, m
Îą Distance?
β House B
CHAPTER 10
SOLUTION OF TRIANGLES
Learning Outcomes: At the end of the lesson, you should be able to: i. Verify the sine rule ii. Use sine rule to find unknown sides or angles of a triangle iii. Find the unknown sides and angles of a triangle involving ambiguous case
Activity 1 C sin A = b
sin B = h=
h
A
h=
a
B
D c
h b b sin A h a a sin B
Equating 1 and 2 b sin A = a sin B a sin A
b = sin B
1
2
Generally‌ The sine rule is written as a sin A
!! ! k Thin
b = sin B
=
c sin C
Can the sine rule be written as sin A a
sin B = b
=
sin C c
?
Sine rule can be used to solve a triangle if given either:
Two angles and one side
x cm ι θ
Two sides and the non-included angle
x cm θ y cm
Example 1: In the triangle KLM as shown, KL = 7.4 cm, KM = 12.8 cm and L = 125ยบ 25'. Find the angle x. K
12.8 cm 7.4 cm 125ยบ 25' L
x
M
Solution: Using the sine rule, sin x 7.4
=
sin x =
sin 125ยบ 25' 12.8 sin 125ยบ 25' 12.8
x 7.4
= 0.4711 Hence, x = 28.1089ยบ x = 28ยบ 7' x is an acute angle
Let’s Try.. In the triangle PQR as shown, PQ = 6.8 cm, and R = 42º. Find the length of PR.
Q = 60º 40'
P 6.8 cm Q
60º 40' 42º R
Solution: Using the sine rule, PR sin 60ยบ 40' Hence,
=
PR =
6.8 sin 42ยบ 6.8 sin 42ยบ
= 8.859 cm
x sin 60ยบ 40'
Activity 2 1. Draw a triangle ABC with the information given: A = 35ยบ, AC = 8 cm and BC = 7 cm 2. In the same triangle, draw another triangle with the same measurement. 3. Can you figure which point gives you different angle for the two triangle that you had drawn?
Based on the information given, two distinct triangles can be form: Δ ABC and Δ AB'C. C 7 cm
8 cm 35º
B
B'
A
The Ambiguous Case of a Triangle An ambiguous case may occur when two
sides (where a<b) and a non-included angle θ are given.
a
b a θ
Example 2: In a triangle PQR, PR = 7cm, QR = 4 cm and Find the probable values of Q. Solution: Draw the probable
PQR
7 cm
20º P
β Q'
P = 20º.
R
4 cm
α Q
Calculations: We know that the sine rule is;
a sin A
=
b sin B
Substitute all the values given, 7 sin Q
=
sin Q =
4 sin 20 7 sin 20
= 0.5985
4 Q = 36º 46' or 143º 14'
Two probable values for
Q are 36º 46' or 143º 14‘.
Try This! Based on the triangle ABC that you had drawn before, find the probable values of B.
C 7 cm
8 cm 35ยบ
B
B'
A
Solution: Using the sine rule, sin
ABC
sin 35º = 8 7 8(sin 35º) sin ABC = 7 8(0.5736) = 7 = 0.6555 ABC = 40.9608º
Hence,
ABC = 40º 58' or (180º - 40º 58' ) = 40º 58' or 139º 2'
Explore Mindâ&#x20AC;Ś. Instructions: 1. Answer the worksheet individually in two 2. 3. 4. 5.
minutes. Discuss the solution in groups. Group leader choose the best answer and write it on manila card. Display and present your group solution in front of the class. The best group will receives coupon.
Question 1(a) Given that in the triangle PQR, PQ = 10 cm, PR = 6 cm and Q = 35ยบ. Determine the length of QR and R.
10 cm
6c m
P
6 cm
35ยบ Q
R1
R2
Solution: Using the sine rule, 10 sin R sin R
= =
6 sin 35º sin 35º 6
x 10
= 0.9560 R R
= 72.9404º = 72º 56' or (180º - 72º 56' ) = 72º 56' or 107º 4'
Therefore,
R1 = 107º 4' and
R2 = 72º 56'
Continue… For triangle PQR1, P
= 180º - 107º 4' - 35º = 37º 56'
QR1 = sin 37º 56' QR1 =
6 sin 35º 6 x sin 37º 56' sin 35º
QR1 = 6.431 cm
Continue… For triangle PQR2, P
= 180º - 72º 56' - 35º = 72º 4'
QR2 sin 72º 4'
6 = sin 35º 6 x sin 72º 4' QR2 = sin 35º = 9.952 cm
Hence,
QR = 6.431 cm and
R = 107º 4' or
QR = 9.952 cm and
R = 72º 56'
Question 1(b) Given that in the triangle ABC, AB = 12 cm, AC = 8 cm and B = 40ยบ. Determine the length of BC and C.
12 cm
8c m
A
8 cm
40ยบ B
C1
C2
Solution: Using the sine rule, 12 sin C sin C
= =
8 sin 40º sin 40º
x 12
C
8 = 0.9642 = 74.6186º
C
= 74º 37' or (180º - 74º 37' ) = 74º 37' or 105º 23'
Therefore,
C1 = 105º 23' and
C2 = 74º 37'
Continue… For triangle ABC1, A
= 180º - 105º 23' - 40º = 34º 37'
BC1 = sin 34º 37' BC1 =
8 sin 40º 8 x sin 34º 37' sin 40º
= 6.96 cm
Continue… For triangle ABC2, A
= 180º - 74º 37' - 40º
= 65º 23' BC2 8 = sin 65º 23' sin 40º 8 x sin 65º 23' BC2 = sin 40º = 11.28 cm Hence,
BC = 6.96 cm and BC = 11.28 cm and
C = 105º 23' or C = 74º 37'
SUMMARY SINE RULE
a sin A
b = sin B
Can be used to solve a triangle if given either: • Two angles and one side • Two sides and the non-included angle
=
c sin C
Ambiguous case of triangle occur when both condition are given: • Two sides ( where a<b) • Non-included angle θ