Law of Sines

General Comments

We learned to solve right triangles in chapter 4. We will start this chapter by learning to solve oblique triangles (non-right triangles).

Please note that angles are Capital letters and the side opposite is the same letter in lower case.

A

C

B

ab

A B

C

ab

c

c

What we already know

• The interior angles total 180.

• We can’t use the Pythagorean Theorem. Why not?

• For later, area = ½ bhA B

C

ab

c

Playing with a the triangle

AB

C

ab

c

Let’s drop an altitude and call it h.

h

If we think of h as being opposite to both A and B, then

sin sinh h

A and Bb a

Let’s solve both for h.

sin sinh b A and h a B

This meanssin sin and dividing by .

sinA sin

a

b A a B ab

B

b

A B

C

ab

c

If I were to drop an altitude to side a, I could come up with

sin sinB C

b c

Putting it all together gives us the Law of Sines.

sin sin sinA B C

a b c

You can also use it upside-down. sin sin sin

a b c

A B C

What good is it?

The Law of Sines can be used to solve the following types of oblique triangles

•Triangles with 2 known angles and 1 side (AAS or ASA)

•Triangles with 2 known sides and 1 angle opposite one of the sides (SSA)

With these types of triangles, you will almost always have enough information/data to fill out one of the fractions.

sin sin sinA B C

a b c

General Process

1. Except for the ASA triangle, you will always have enough information for 1 full fraction and half of another. Start with that to find a fourth piece of data.

2. Once you know 2 angles, you can subtract from 180 to find the 3rd.

3. To avoid rounding error, use given data instead of computed data whenever possible.

Example 1 (AAS)

A B

C

ab

c

45 , 50 , 30Let A B a

Once I have 2 angles, I can find the missing angle by subtracting from 180. C=180 – 45 – 50 = 85

45 50

=3085 I’m given both pieces for

sinA/a and part of sinB/b, so we start there.sin 45 sin50

30 b

Cross multiply and divide to get

sin 45 30sin50

30sin50

sin 45

b

b

45 50

=3085

A B

C

ab

c

30sin50

sin 45b

Using a calculator,

b 32.5

b 32.5

We’ll repeat the process to find side c.

Remember to avoid rounded values when computing.

sin sin

sin 45 sin85

30

A C

a c

c

Cross multiply and divide to get

sin 45 30sin85

30sin85

sin 45

c

c

Using a calculator,

42.3c

c 42.3

We’re done when we know all 3 sides and all 3 angles.

Example 2 ASA

A B

C

ab

c

Let A = 35, B = 10, and c = 45

35 10

45

Since we can’t start one of the fractions, we’ll start by finding C.

C = 180 – 35 – 10 = 135

135

Since the angles were exact, this isn’t a rounded value. We use sinC/c as our starting fraction.

sin sin sin sinC A C Band

c a c b

sin135 sin 35

45 a

Cross multiply and divide

sin135 45sin 35

45sin 35

sin135

a

a

Using your calculator

36.5a

36.5

sin135 sin10

45 b

sin135 45sin10

45sin10

sin135

b

b

11.1b

11.1

Example 3 SSA

A B

C

ab

c

Let A = 40, b = 10, and a = 7

We have enough information to use sinA/a as our starting fraction and can go to work on B

sin sin

sin 40 sin

7 10

A B

a bB

We cross multiply and divide to get

10sin 40sin

7B

We have to get rid of the sin function to find the answer for angle B.

To do this, we apply the inverse operation.

1

1

1 10sin 40sin

7s

10sin 40sin

7

in sinB

B

With a calculator

66.7B

40

10 7

66.7

40

10 7

66.7

A B

C

ab

c

We need something to start the fraction sinC/c. Since it is critical that the angles total 180, we will find C by subtracting from 180.

C=180 – 40 – 66.7 = 73.3

73.3

Now we use our starting fraction, sinA/a with sinC/c.

sin 40 sin 73.3

7 c

Cross multiply and divide to finish.

sin 40 7sin 73.3

7sin 73.310.4

sin 40

c

c

10.4

A Pain in the Angle Side SideLet’s consider the case where we have an angle, an adjacent side, and an opposite side. For example, I have angle A, side b, and side a.

Sometimes a is too short to reach. You attempt to work this problem like example 3. Your calculator will give you an error message and catch the error.A

b a

Sometimes a is just right. It reaches with a right angle. You work this problem like example 3 and will never know there might have been a problem.

A

ba

Sometimes a is so long it only reaches one way. This problem also works just like example 3. You’ll never know this might have been difficult.A

b aNo triangle on this side

Sometimes a is just the right length that it can form 2 different triangles. Following example 3 solves the outer triangle. You have to be on the look-out to catch the second triangle.

aa

A

b

If you don’t get an error and a < b, then there will be either a right angle or 2 two triangles.

Contemplating Triangles

A

baa

Outer triangle

Inner triangle

C

c

BBB’

c’

Solving like example 3 will give you the purple (outer) triangle.

Since the 2 red lines (a) are equal, the 2 base angles (B) are also equal.

Since B and B’ form a straight line, B’ = 180 – B.

Use B’ to compute C’. Subtract A and B’ from 180 to find C’

C’

Another Perspective on Starting the Second Triangle

a a

If the 2 sides are equal, the 2 angles are equal.

BB

Since supplementary angles total 180, B’ = 180 – B.

B’ = 180 - B

A

b

c’

C’

Since A, b, and a are given, once you know B’ you can finish the triangle on the left.

An Example of a Pain in the Angle Side Side

Let A = 40, b = 10, and a = 9. Angle side side

A B

C

40

b = 10 a = 9

We will proceed like example 3.

We have enough information for A and a start for B

sin sin

sin 40 sin

9 10

A B

a bB

9sin 10sin 40

10sin 40sin

9

B

B

Cross multiply and divide to get

c

1 1

1

10sin 40sin sin sin

9

10sin 40sin 45.6

9

B

B

To get to angle B, you must unlock sin using the inverse.

40

b = 10

A B

C

a = 9

45.6

Once you know 2 angles, find the third by subtracting from 180.

C = 180 – 40 – 45.6 = 94.4

94.4

c

We’re ready to look for side c. Remember to use the starter fraction (A)

sin sin

sin 40 sin 94.4

9

A C

a c

c

Cross multiply and divide. Then use a calculator.

sin 40 9sin 94.4

9sin 94.414.0

sin 40

c

c

=14.0

Finding the Second Triangle

A

40

b= 10a = 9

B = 45.6B’

Start by finding B’ = 180 - B

B’ = 180 – 45.6 = 134.4

Now solve this triangle.

A B’

C’

b= 10

a = 9

40 134.4

c’

A B’

C’

b= 10

a = 9

40 134.4

c

Start by finding C’ = 180 – 40 – 134.4

C’ = 5.65.6

Finish by finding side c, using your starter fraction (A)

sin sin

sin 40 sin5.6

9sin 40 9sin5.6

9sin5.61.4

sin 40

A C

a c

cc

c

=1.4

A New Way to Find AreaWe all know that A = ½ bh.

And a few slides back we found this.

sin sinh b A and h a B A B

C

ab

c

It looks like the base is c.

Making a little substitution gives me

1 1sin sin

2 2Area bc A ac B

If we drop the altitude in a different direction, we can add

1sin

2Area ab C

It’s always the 2 sides and the included angle.

An Area ExampleFind the area of a triangle with side a = 10, side b = 12, and angle C = 40.

1. Choose the appropriate area formula.

In this case we choose Area = ½ ab sinC.

2. Fill in the blanks.

Area = ½ (10)(12)sin40

3. Plug it into your calculator

Area = 38.6 square units