9.3 The Law of Sines9.3 The Law of Sines

9.3/9.4 Laws of Sines and Cosines

Objectives:1. Solve non-right triangles.

Vocabulary: Law of Sines, Law of Cosines

Derivation of the Law of Sines

Cc

Aa

AcCa

Cahah

C

Achch

A

sinsinsinsin

sinorsin

,sinorsin

sinAsinC sinAsinC

Derivation of the Law of Sines

Is the same true if we start with an obtuse triangle?.

sinAsinC sinAsinC

When A is obtuse, then noting that sin( – A) = sinA

– A

sin( ) or sin( )

sin ,

sin or sin

sin sinsin sin

hA h c Ac

c AhC h a Ca

a ca C c AA C

sin sin sinA B C

a b c

A B

C

ab

c

The Law of Sines

Alternative forms are sometimes convenient to use:

.sinsinsin C

c

B

b

A

a

Using the Law of Sines in an Application (ASA)

Two stations are on an east-west line 110 miles apart. A forest fire is located on a bearing of N 42oE from the western station at A and a bearing of N 15oE from the eastern station at B. How far is the fire from the western station?

A = 90o – 42o = 48o

B = 90o + 15o = 105o

C = 180o – 105o – 48o = 27o

Using the law of sines to find b gives

o o

110 , then 234 miles.sin105 sin 27

b b

First find the measure of :C 180 110 20 50C

Then by the law of sines:

sin110 sin 20 sin50

25a b

sin110 sin50

25a

25sin110 sin 50a

25sin110

sin50a

30.7

sin 20 sin50

25b

25sin 20 sin 50b

25sin 20

sin 50b

11.2

So, 30.7 m and 11.2 m.BC AC

A civil engineer wants to determine the distances from points A and B to an inaccessible point C, as shown. From direct measurement the engineer knows that AB = 25 m, A = 110o, and B = 20o. Find AC and BC.