Law of Sines In Section 14-3 you studied techniques for solving - - PowerPoint PPT Presentation

Section 14-4 Law of Sines

In Section 14-3 you studied techniques for solving right triangles. In this section you will learn techniques for solving non-right triangles or oblique triangles.

A C B b c a

To solve an oblique triangle, you need to know the measure of at least one side and any two other parts of the triangle.

In this section we will solve oblique triangles knowing the following Two angles and one side (ASA or AAS)

Law of Sines

If is a triangle with sides , , and , then ABC a b c sin sin sin .   A B C a b c

Example 1

Find the remaining angle and sides.

102.3° 28.7° 27.4 ft c C B a A

To find A subtract B and C from 180°. A = 180 – B – C A = 180 – 28.7° – 102.3° A = 49.0°

sin49 sin28.7 27.4  a Substitute 27.4 for b solve for a. 27.4sin49 sin28.7  a 43.06 ft  sin sin sin   A B C a b c

sin102.3 sin28.7 27.4  c Substitute 27.4 for b solve for c. 27.4sin102.3 sin28.7  c 55.75 ft  sin sin sin   A B C a b c

Example 2

Find the remaining angle and sides.

30° 45° 32 ft c C B b A

To find C subtract A and B from 180°. C = 180 – A – B C = 180 – 30° – 45° C = 105°

sin45 sin30 32  b Substitute 32 for a solve for b. 32sin45 sin30  b 45.25 ft  sin sin sin   A B C a b c

sin105 sin30 32  c Substitute 32 for a solve for c. 32sin105 sin30  c 61.82 ft  sin sin sin   A B C a b c

Example 3

A surveyor locates points A and B at the same elevation and 3950 ft.

• apart. At A, the angle of elevation

to the summit of the mountain is 18°. At B, the angle of elevation is 31°. (a) Find the distance from B to the summit. (b) Find the height of the mountain.

A B 3950 ft. 18° 31° x C

To find x we need to find the mACB. We must first find the mABC. mABC = 180° − 31° = 149°

mACB = 180° − 18° − 149° = 13° Now we use the law of sines to solve for x.

A B 3950 ft. 18° 31° x C 149°

A B 3950 ft. 18° 31° x C 13°

sin13 sin18 3950    x 3950sin18 sin13    x 5426 ft. 

A B 3950 ft. 18° 31° 5426 C 13° h

Now use the trig ratios from Section 14-3 to the height. sin31 5426   h 5426sin31   h 2795 ft. 

Area of an Oblique Triangle The area of any triangle is one-half the product of the lengths of two sides and the sine of the included

• angle. That is,

1 Area sin 2  bc A 1 Area sin 2  ab C A C B b c a 1 Area sin 2  ac B

Example 4

Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102°.

Let a = 90 meters b = 52 meters C = 102°

90 meters 52 meters 102°

   

1 Area 90 52 sin102 2 

2

Area 2288.87 m 