Warm-Up 10/15
1.
H
PSAT Tomorrow 10/16, you will need to bring
your own calculator.
Rigor:
You will learn how to use the Law of
Sines and Cosines to solve right
triangle problems.
Relevance:
You will be able to use the Law of Sines
and Cosines to solve real world
problems.
4-7b Law of Sines and
the Law of Cosines
Cases:
1
2
3
4
ASA or SAA
SSA
SAS
SSS
LAW OF SINES
LAW OF COSINES
Example 4: Find two triangles for which A = 43°, a = 25, b = 28.
Round side lengths to the nearest tenth and angle measures to
the nearest degree.
A is acute, and h = 28sin 43° ≈ 19.1
Since a < b and a > h there are two different triangles.
Solution 1: Find the acute  B.
sin 𝐵 𝑠𝑖𝑛 43°
sin 43° 𝑠𝑖𝑛 87°
=
≈
28
25
25
𝑐
28𝑠𝑖𝑛 43°
𝑐 sin 43° ≈ 25𝑠𝑖𝑛 87°
sin 𝐵 =
25
25𝑠𝑖𝑛 87°
28𝑠𝑖𝑛
43°
𝑐≈
𝐵 = 𝑠𝑖𝑛−1
sin 43°
25
𝐵 ≈ 50°
𝑐 ≈ 36.6
𝐶 ≈ 180° − 43° − 50° ≈ 87°
B ≈ 50°, C ≈ 87°, and c ≈ 36.6
Example 4: Find two triangles for which A = 43°, a = 25, b = 28.
Round side lengths to the nearest tenth and angle measures to
the nearest degree.
Solution 2: Find the obtuse  B'.
CB'B =  , so  B'≈ 180°– 50° ≈ 130°
B
 C≈ 180°– 43 °– 130° ≈ 7°
sin 7° 𝑠𝑖𝑛 43°
≈
𝑐
25
𝑐 sin 43° ≈ 25𝑠𝑖𝑛 7°
25𝑠𝑖𝑛 7°
𝑐≈
sin 43°
𝑐 ≈ 4.5
B′ ≈ 130°, C ≈ 7°, and c ≈ 4.5
Law of Cosines:
a  b  c  2bc cos A
2
2
2
b  a  c  2ac cos B
2
2
2
c  b  a  2ab cos C
2
2
2
Example 5: Solve ∆ABC. Round side lengths to the nearest tenth
and angle measures to the nearest degree.
Given a = 6, b =24, c = 20 find angle A.
𝑎2 = 𝑏 2 +𝑐 2 −2𝑏𝑐 cos 𝐴
62 = 242 +202 −2 24 20 cos 𝐴
36 = 576 + 400 − 960 cos 𝐴
36 = 976 − 960 cos 𝐴
−940 = −960 cos 𝐴
940
= cos 𝐴
960
940
−1
𝑐𝑜𝑠
=𝐴
960
𝐴 ≈ 11.7°
Example 6: Solve ∆ABC. Round side lengths to the nearest tenth
and angle measures to the nearest degree.
𝑐 2 = 𝑎2 +𝑏 2 −2𝑎𝑏 cos 𝐶
𝑐 2 = 52 +82 −2 5 8 cos 65°
𝑐 2 ≈ 55.19
𝑐 ≈ 7.4
sin 𝐴 𝑠𝑖𝑛 65°
=
5
7.4
5𝑠𝑖𝑛 65°
sin 𝐴 =
7.4
5𝑠𝑖𝑛 65°
−1
𝐴 = 𝑠𝑖𝑛
7.4
A ≈ 38°
𝐵 ≈ 180° − 65° − 38° ≈ 77°
A ≈ 38°, B ≈ 77°, and c ≈ 7.4
Law of Cosines:
b c a
cos A 
2bc
2
2
2
a c b
cos B 
2ac
2
2
2
a b c
cos C 
2ab
2
2
2
Heron’s Formula: SSS
Area  s  s  a  s  b  s  c 
1
s  a  b  c
2
Example 7: Find the area of ∆XYZ.
𝑥 = 45 𝑖𝑛. 𝑦 = 51 𝑖𝑛. 𝑧 = 38 𝑖𝑛.
1
𝑠 = 𝑥+𝑦+𝑧
2
1
𝑠 = 45 + 51 + 38
2
𝑠 = 67
𝐴𝑟𝑒𝑎 =
𝑠 𝑠−𝑥 𝑠−𝑦 𝑠−𝑧
𝐴𝑟𝑒𝑎 =
67 67 − 45 67 − 51 67 − 38
𝐴𝑟𝑒𝑎 = 683936
𝐴𝑟𝑒𝑎 ≈ 827 𝑖𝑛2
Example 8: Find the area of ∆GHJ to the nearest tenth.
𝑔 = 7𝑐𝑚
ℎ = 10 𝑐𝑚
𝐽 = 108°
𝐴𝑟𝑒𝑎 =
1
𝑔ℎ sin 𝐽
2
𝐴𝑟𝑒𝑎 =
1
7 10 sin 108°
2
𝐴𝑟𝑒𝑎 ≈ 33.3 𝑐𝑚2
−1
math!
4-7a Assignment: TX p298, 2-24 even
Unit 4 Test
Monday 10/21
−1
math!
4-7a Assignment: TX p298, 2-24 even
Assignment for Wednesday:
4-7b Assignment: TX p298-299, 26-50 even
Unit 4 Test
Monday 10/21