10.5B Law of Sines
Objectives:
G.SRT.10: Prove the Laws of Sines and Cosines and use them to solve problems.
G.SRT.11: Understand and apply the Law of Sines and the Law of Cosines to find unknown
measurements in right triangles.
For the Board: You will be able to determine the area of a triangle given side-angle-side information
and use the Law of Sines to find the side lengths and angle measures of a triangle.
Anticipatory Set:
The area of ΔABC has 3 different formulas which all must result in the same value.
½ bc sin A = ½ ac sin B = ½ ab sin C
B
The ½ ‘s cancel out so bc sin A = ac sin B = ab sin C
a
Divide through by abc: bc sin A = ac sin B = ab sin C
c
abc
abc
abc
This results in sin A = sin B = sin C
A
b
a
b
c
This is known as the Law of Sines.
C
Instruction:
The Law of Sines allows you to solve a triangle as long as you know either of the following:
1. Two angle measures and any side length: AAS or ASA.
2. Two side lengths and the measure of an angle that is not between them: SSA.
Open the book to page 723 and read example 2.
Example: Solve each triangle. Round to the nearest tenth.
A. AAS
33° + m<E + 28° = 180° so m<E = 119°
sin 33 sin119 sin 28
E


d
e
15
d
15
e sin 28 = 15 sin 119 so
15 sin 119
28°
F
e=
= 27.9
33°
D
e
sin 28
d sin 28 = 15 sin 33 so
15 sin 33
d=
= 17.4
sin 28
B. ASA
P
q
r
Q
36°
39°
10
R
36° + m<P + 39° = 180° so m<P = 105°
sin 36 sin105 sin 39


q
10
r
q sin 105 = 10 sin 36 so
10 sin 36
q=
= 6.1
sin 105
r sin 105 = 10 sin 39 so
10 sin 39
r=
= 6.5
sin 105
White Board Activity:
Practice: Solve each triangle. Round to the nearest tenth.
A.
B.
J
k
H
h
2
107°
42°
12
1.5
K
42° + 107° + m<K = 180°
m<K = 31°
sin42 sin107 sin31


h
12
k
h sin 107 = 12 sin 42
12 sin 42
h=
= 8.4
sin 107
k sin 107 = 12 sin 31
12 sin 31
k=
= 6.5
sin 107
Assessment:
Question student pairs.
Independent Practice:
Text: pgs. 726 – 727 prob. 4 – 9, 17 – 19, 25 - 29.
For a Grade:
Text: pgs. 726 – 727 prob. 6, 18, 24.
P
56°
M
106°
p
m
56° + 106° + m<N = 180°
m<N = 18°
sin56 sin106 sin18


p
m
1.5
p sin 18 = 1.5 sin 56
1.5 sin 56
p=
= 4.0
sin 18
m sin 28 = 1.5 sin 106
1.5 sin 106
m=
= 4.7
sin 18
N