6.1 Law of Sines

If ABC is an oblique triangle with sides a, b, and c, then

C

c

B

b

A

a

sinsinsin==

A B

C

c

b a

Ex. 1 Given a triangle with C = 102.3o, and B = 28.7o, and b = 27.4 feet, find the remaining angle and sides. Given: AAS

A = 180 - B - C

A = 49o

A B

C

c

ab = 27.4’

28.7o

102.3o

°=

° 49sin7.28sin

4.27 a

.06.43 fta ≈

Set up and find a.

Now find side c. .75.55 ftc ≈

For the SSA oblique triangle there are 3 possible situations that can occur:

1. No triangle exists h > a

2. One triangle exists a > b

3. Two distinct triangles exist. h < a < b

Let’s look at the first and third cases.

Case 1: No-solution case - SSA Where h > a

Given: a = 15, b= 25 , and A = 70o

b = 25 a = 15

A

How do we find the height of the triangle?

2570sin

h=°

h

h = 23.49 which is longer than opposite side a.

In other words, h > a which implies no solution.φ

°70

Case 2: Two solutions h < a < b Draw two triangles.

Given: a = 12 m, b = 31 m, and A = 20.5o

20.50

b = 31

BB’

a = 12 a = 12

c’

c

h

A

C’

C

A’

The first thing we need to do after the triangles are drawn is to find the height to see is the triangle even exists. Find the height.

315.20sin

h= h = 10.86 which is less than 12.

Now we can find B, B’, C, C’, c, c’

First, find B.Bsin

31

5.20sin

12=

°B = 64.8o

oB 2.115'=∴Now, find C and C’

C = 94.7o

C’ = 44.3o

Now, use law of sines to find c and c’. c = 34.15

c’ = 23.93

The Area of an Oblique Triangle

A = (1/2) bh

Ex. Find the area of the given triangle.

102o

a = 90 m

b = 52 m

Draw in the height and find it by finding the supplement to 102o.o.

h

78o 5278sin

ho =

h = 50.86 m

27.2288)86.50)(90(2

1mmmA ≈=