GEOMETRY
MODULE 2 LESSON 32
USING TRIGONOMETRY TO FIND SIDE LENGTHS OF AN ACUTE
TRIANGLE
OPENING EXERCISE
Complete both parts of the Opening Exercise in your workbook.
sin 60 =
𝑒=
5
𝑒
5
10
=
sin 60 √3
tan 30 =
𝑑
5
𝑑 = 5 tan 30 =
5
√3
We do not have enough information to solve the missing
lengths. We cannot assume this is a right triangle.
DISCUSSION
There are two facts in trigonometry that help us find unknown measurements in triangles that are not
right triangles. These facts can be used for both acute and obtuse triangles. (Today, we will study
acute triangles only.)
LAW OF SINES
sin ∠𝐴 sin ∠𝐵 sin ∠𝐶
=
=
𝑎
𝑏
𝑐
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Example 1 (Workbook): A surveyor needs to determine the
distance between two points A and B that lie on opposite banks of a
river. A point C is chosen 160 meters from point A, on the same
side of the river as A. The measures of ∠𝐵𝐴𝐶 and ∠𝐴𝐶𝐵 are 41°
and 55°, respectively. Approximate the distance from A to B to the
nearest meter.

Use
sin ∠𝐴
𝑎
=
sin ∠𝐵
𝑏
=
sin ∠𝐶
𝑐
and substitute the known values.
sin 41 sin 84 sin 55
=
=
𝑎
160
𝑐

Which ratio is relevant to finding the distance from A to B? Note: side𝐴𝐵 = 𝑐
sin 84 sin 55
=
160
𝑐

Solve by cross multiply.
𝑐 sin 84 = 160 sin 55
𝑐 =
160 sin 55
sin 84
𝑐 = 132 𝑚𝑒𝑡𝑒𝑟𝑠
ON YOUR OWN
Complete Exercise 1 in your workbook.
sin 30 sin ∠𝐵
=
12
10
12 sin ∠𝐵 = 10 sin 30
sin ∠𝐵 =
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10 sin 30
1
5
= 10 ( ) ÷ 12 =
12
2
12
2
PRACTICE
Consider Exercise 2 in your workbook.
A car is moving toward a tunnel carved out of the base of a hill. As
the accompanying diagram shows, the top of the hill, H, is sighted
from two location, A and B. The distance between A and B is 250 ft.
What is the height, h, of the hill to the nearest foot?
sin 15 sin 30
=
250
𝑥
𝑥 sin 15 = 250 sin 30
250 sin 30 250(. 5)
125
𝑥=
=
=
sin 15
sin 15
sin 15
𝑥 = 483 𝑓𝑡
ℎ√2 = 𝑥
ℎ√2 = 483
ℎ=
483
√2
ℎ = 342 𝑓𝑡
LAW OF COSINES
𝑐 2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 cos ∠𝐶
Note: There are other possible arrangements for the law of cosines

𝑎2 = 𝑏 2 + 𝑐 2 − 2𝑏𝑐 cos ∠𝐴

𝑏 2 = 𝑎2 + 𝑐 2 − 2𝑎𝑐 cos ∠𝐵

To solve for a missing side, we must know two side lengths and the measure of the included
angle.
PRACTICE
Find side d to the nearest tenth.
𝑑 2 = 𝑒 2 + 𝑓 2 − 2𝑒𝑓 cos ∠𝐷
𝑑 2 = 62 + 92 − 2(6)(9) cos 65
𝑑 2 = 36 + 81 − 108 cos 65
𝑑 2 = 117 − 108 cos 65
𝑑 = √117 − 108 cos 65
𝑑 ≈ 8.4
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Example 2 (Workbook): Our friend the surveyor from Example 1 is doing
some further work. He has already found the distance between points A and
B (from Example 1). Now he wants to locate a point D that is equidistant
from both A and B and on the same side of the river as A. He has his
assistant mark the point D so that ∠𝐴𝐵𝐷 and ∠𝐵𝐴𝐷 both measure 75°.
What is the distance between D and A to the nearest meter?

Label side AD as b and side BD as a.

Side AB was found earlier to be 132.

We are trying to find b.
𝑏 2 = 𝑎2 + 𝑑 2 − 2𝑎𝑑 cos ∠𝐵
Since this is an isosceles triangle, 𝑎 = 𝑏
𝑏 2 = 𝑏 2 + 1322 − 2(𝑏)(132) cos 75
𝑏 2 = 𝑏 2 + 17424 − 264(𝑏) cos 75
0 = 17424 − 264(𝑏) cos 75
264(𝑏) cos 75 = 17424
𝑏=
17424
264 cos 75
𝑏 = 255𝑚
ON YOUR OWN
Attempt Exercise 3 in your workbook.
In the parallelogram, ∠𝐶 = 100°, therefore ∠𝐷 = 80°.
𝑑2 = 𝑎2 + 𝑐 2 − 2𝑎𝑐 cos ∠𝐷
𝑑2 = 262 + 442 − 2(26)(44) cos 80
𝑑 = √262 + 442 − 2(26)(44) cos 80
𝑑 = 47𝑚𝑚
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SUMMARY

Law of Sines:
sin ∠𝐴
𝑎
=
sin ∠𝐵
𝑏
=
sin ∠𝐶
𝑐

Law of Cosines: 𝑐 = 𝑎 + 𝑏 − 2𝑎𝑏 cos ∠𝐶

We apply the laws of sines and cosines when we do not have right triangles to work with. These
2
2
2
laws can be used for both acute and obtuse triangles.
HOMEWORK
Problem Set Module 2 Lesson 32, page 240
#1 and #6. Round answers to the nearest tenth.
Show all work in an organized and linear manner.
DUE: Tuesday, Jan 30, 2017
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