PRE-CALCULUS LESSON 6.1LAW OF SINES

Spring 2011

Question! How to measure the depth?

http://www.mcs.uvawise.edu/dbl5h/resources/latex_examples/files/triglawsines.pdf

Use the Law of Sines to solve oblique triangles (AAS or ASA).

Use the Law of Sines to solve oblique triangles (SSA).

Find areas of oblique triangles. Use the Law of Sines to model and solve

real-life problems.

At the end of this lesson you should be able to

Introduction

1. To solve an oblique triangle, we need to be given at least one side and then any other two parts of the triangle.

2. The sum of the interior angles is 180°.

3. Why can’t we use the Pythagorean Theorem?

4. State the Law of Sines.

5. Area = ½ base*height

6. Area= ½*a*b*sineC

A B

C

ab

c

Introduction (continued)

7. Draw a diagram to represent the information. ( Do not solve this problem.)

A triangular plot of land has interior angles A=95° and C=68°. If the side between these angles is 115 yards long, what are the lengths of the other two sides?

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

Taking reciprocals, we have

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

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 sin 50

30 b

Cross multiply and divide to get

sin 45 30sin 50

30sin 50

sin 45

b

b

45° 50°

=3085°

AB

C

ab

c

30sin 50

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

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 triangular plot of land has interior angles A=95° and C=68°. If the side between these angles is 115 yards long, what are the lengths of the other two sides?

The Ambiguous Case (SSA)

Three possible situations1. No such triangle exists.2. Only one such triangle exists.3. Two distinct triangles can satisfy the conditions.

Example 3(SSA)

Use the Law of Sines to solve the triangle.A = 26, a = 21 inches, b = 5 inches

A

b = 5 ina = 21 in26

a > b : One Triangle

B

b

A

a

sinsin

21

26sin5sin

sin

5

26sin

21

B

B1 5 sin 26

sin ( )21

5.99

B

B

B

C

Example 3(SSA)

Use the Law of Sines to solve the triangle.

A = 26, a = 21 inches, b = 5 inches

A

b = 5 ina = 21 in26

a > b : One Triangle

B

C 01.148)99.526(180C

38.2526sin

01.148sin2101.148sin26sin

21

c

c5.99°

Example 4(SSA)

Use the Law of Sines to solve the triangle.A = 76, a = 18 inches, b = 20 inches

sin sina b

A B

18 2076sin sin B

sin 1.078B

There is no angle whose sine is 1.078.

There is no triangle satisfying the given conditions.

AB

b = 20 ina = 18 in

76

a < h:None

4.19

76sin20

sin

h

h

Abh

h

Example 5(SSA)

Let A = 40°, b = 10, and a = 9.

AB

C

40°

b = 10 a = 9We have enough information for

sin sin

sin 40 sin

9 10

A B

a bB

9sin 10sin 40

10sin 40sin

9

B

B

Cross multiply and divide

c

h < a < b: Two

4.6

40sin10

sin

h

h

Abh

h=6.4

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.

sin sin

sin 40 sin 94.4

9

A C

a c

c

sin 40 9sin 94.4

9sin 94.414.0

sin 40

c

c

=14.0

Example 5Finding the Second Triangle

A40°

b= 10 a = 9

B = 45.6°B’

Start by finding B’ = 180 - B

B’ = 180 – 45.6 = 134.4°

Now solve this triangle.

A B’

C’

b= 10a = 9

40° 134.4°

c’

Let A = 40°, b = 10, and a = 9.

A B’

C’

b= 10

a = 9

40° 134.4°

c

Next, find C’ = 180 – (40 + 134.4)

C’ = 5.6°5.6°

sin sin

sin 40 sin 5.6

9sin 40 9sin 5.6

9sin 5.61.4

sin 40

A C

a c

cc

c

=1.4

A New Way to Find Area

We 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

1 1 1Area sin sin sin2 2 2

bc A ab C ac B

h

Area = ½*product of two given sides * sine of the included angle

Example 6Finding the area of the triangle

Find the area of a triangle with side a = 10, side b = 12, and angle C = 40°.

22

Two fire ranger towers lie on the east-west line and are 5miles apart. There is a fire with a bearing of N27°E from tower 1 and N32°W from tower 2. How far is the fire from tower 1?The angle at the fire is 180° - (63° + 58°) = 59°.

1 25 mi

63° 58°

xN

S

N

S

Example 7 Application

Example 8 Application

http://www.mcs.uvawise.edu/dbl5h/resources/latex_examples/files/triglawsines.pdf

24

p.398-400 #s 2-38, even

Practice