13.5 Apply the Law of Sines

 

Goal · Solve triangles that have no right angle.

 

Your Notes

 

VOCABULARY

Law of sines The law of sines can be used to solve triangles when two angles and the length of any side are known (AAS or ASA cases), or when the lengths of two sides and an angle opposite one of the two sides are known (SSA case).

 

LAW OF SINES

The law of sines can be written in either of the following forms for ΔABC with sides of lengths a, b, and c.

 

 

 

 


Example 1

Solve a triangle for the AAS or ASA case

 

Solve ΔABC with A = 28°, B = 102°, and a = 8.

 

Solution

First find the angle: C = 180° - 102° - 28° = 50°.

 

 

 

Write equations

b =

Solve for each variable.

b » 16.7

 

Use a calculator.

 

 

Write equations.

c =

Solve for each variable.

c » 13.1

Use a calculator.

 

In ΔABC, C = 50° , b »16.7 , and c » _13.1_ .

Your Notes

 

POSSIBLE TRIANGLES IN THE SSA CASE

Consider a triangle in which you are given a, b, and A. By fixing side b and angle A, you can sketch the possible positions of side a to figure out how many triangles can be formed. In the diagrams below, note that h = ib sin A.

 

A is obtuse.

A is acute.

a £ b

_No triangle_

 

h > a.                                     h = a

_No triangle_                         _One triangle_

 

a > b

_One triangle_

h < a < b                                a > b

_Two triangles_                      _One triangle_

 

 

Example 2

Solve the SSA case with one solution

 

Solve ΔABC with A = 94°, a = 18, and c = 13.

Make a sketch. Because A is _obtuse_ and a is longer than c, _only one triangle_ can be formed. Use the law of sines to find C.

 

Law of sines

sin C =                  » 0.7205

Multiply each side by 13 .

C » 46.1°

Use inverse sine function.

You then know that B » _180° - 94° - 46.1°_ = _39.9°_

 

Law of sines

b =

Multiply each side by sin 39.9° .

 

b » _11.6_

Use a calculator.

In ΔABC, C » 46.1° , B » 39.9° , and » 11.6 .

Your Notes

 

Example 3

Examine the SSA case with no solution

 

Solve ΔABC with A = 77°, a = 6.1, and b = 9.

 

Solution

 

 


Begin by drawing a horizontal line. On one end form a 77° angle (A) and draw a segment _9_ units long (          or b). At vertex C, draw a segment _6.1_ units long (a). You can see that a needs to be at least _9 sin 77°_ » _8.8_units long to reach the horizontal side and form a triangle. So, it is _not possible_ to draw the indicated triangle.

 Checkpoint Solve ΔABC.

1.      C = 14º, B = 117º, b = 21

A = 49º, a » 17.8, and c » 5.7

2.      A = 56º, a = 24, b = 16

B » 33.6º, C » 90.4º, c » 28.9

3.      B = 122º, b = 5, a = 8

no solution

Your Notes

 

Example 4

Solve the SSA case with two solutions

 

Solve ΔABC with A = 30°, a = 10, and b = 15.

 

Solution

First make a sketch. Because b sin A = __15 sin 30°__ = _7.5_, and 7.5 < 10 < 15 (h < a < b), two triangles can be formed.

 

Triangle 1

Triangle 2

 

Use the law of sines to find the possible measures of B.

Law of sines

sin B =                  = _0.75_

Evaluate.

 

There are two angles B between 0° and 180° for which sin B = _0.75_. One is acute and the other is obtuse. Use your calculator to find the acute angle:

sin-1 0.75 » _48.6°_.

 

The obtuse angle has _48.6°_ as a reference angle, so its measure

is _180° - 48.6°_ = _131.4°_. Therefore, B » _48.6°_ or B » _131.4°_.

Now find the remaining angle C and side length c for each triangle.

Triangle 1                                  Triangle 2

C » _180° - 30° - 48.6°_                    C » _180° - 30° - 131.4°_

= _101.4°_                                                = _18.6°_

                = ____________                                 = _________

c = ____________                                                    c = ____________

» _19.6_                                              » _6.4_

In Triangle 1, B _48.6°_,                     In Triangle 1, B _131.4°_,

C » _101.4°_, and                             C » _18.6°_, and

c » _19.6_.                                        c » _6.4_.

 

Your Notes

 

AREA OF A TRIANGLE

The area of any triangle is given by one half the product of the lengths of two sides times the sine of their included angle. For AABC shown, there are three ways to calculate the area:

 

Area =

Area =

Area =

 

 

Example 5

Find the area of a triangle

Land A piece of land is bordered by three roads as shown. Find the area of the land.

 

Solution

Area =

Write area formula.

=

Substitute.

»_1.6_

Use a calculator.

The area of the land is about _1.6_ square miles.

 


Checkpoint Complete the following exercises.

4.     Solve ΔABC when A = 35°, a = 11, and b = 14.

B » 46.9°, C » 98.1°, and c » 19.0; or

B » 133.1°, C » 11.9°, and c » 4.0

5.      Suppose the side lengths in Example 5 are 4.6 miles and 2.8 miles. Find the area.

about 6.3 mi2

 

Homework

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