﻿ Law of Cosines Challenge problems and word problems, explained step by step # The Law of Cosines

## Word problems, challenge problems

This page assumes that you have a basic understanding of how to use law of cosines formula, and this page focuses solely on the law of cosines (no law of sines problems on this page)

### Practice Problems

##### Problem 1

Don't let the $$29 ^{\circ }$$ angle fool you.

Remember: you can only use an angle when you are trying to solve for the 3rd side of a triangle! The $$29^ \circ$$ does nothing for the law of cosines. (As an aside, you could use that angle with the law of sines .) You can even cross it out , if you want.

Step 1
Determine the appropriate form of the formula

Since we want to find the value of $$\angle B$$, we need

$$b^2 = a^2 + c^2 -2ac \cdot cos( {\color{red}{B}} ) \\$$

Step 2

Substitute the values into the formula

$$b^2 = a^2 + c^2 -2ac \cdot cos( {\color{red}{B}} ) \\ 12^2 = 8^2 + 16^2 -2\cdot 8 \cdot 16 \cdot cos( {\color{red}{B}} )$$

Step 3

Solve

$$12^2 = 8^2 + 16^2 -2\cdot 8 \cdot 16 \cdot cos( {\color{red}{B}} ) \\ 144= 320 - 256 cos( {\color{red}{B}}) \\ \frac{144 -320}{-256} = cos( {\color{red}{B}}) \\ 0.6875 = cos( {\color{red}{B}}) \\ {\color{red}{B}} = cos^{-1}(0.6875) \\ {\color{red}{B}} = 46.56746344221023 ^ \circ$$

##### Problem 2

Don't let the $$53 ^{\circ}$$ angle fool you.

Remember: the law of cosines requires the use of the included angle. The $$53^ \circ$$ does nothing for you. ( Unless you know the law of sines.) You can even cross it out , if you want.

Step 1

Determine the appropriate form of formula

Since we want to find side c we need
$${\color{red}{c}}^2 = a^2 + b^2 -2ab \cdot cos(C) \\$$

Step 2

Substitute the values into the formula

$${\color{red}{c}}^2 = a^2 + b^2 -2ab \cdot cos( C) \\ {\color{red}{c}}^2 = 14^2 + 13^2 -2\cdot 14 \cdot 13 \cdot cos( 79)$$

Step 3

Solve

$${\color{red}{c}}^2 = 14^2 + 13^2 -2\cdot 14 \cdot 13 \cdot cos( 79) \\ {\color{red}{c}}^2 = 295.5455256829377 \\ {\color{red}{c}} = \sqrt{ 295.5455256829377 } \\ {\color{red}{c}} = 17.191437568828782$$

##### Problem 3

Don't let the $$22 ^{\circ}$$ angle fool you.

Remember: you can only use an angle when you are trying to solve for the 3rd side of a triangle! The $$22^ \circ$$ does nothing for you. ( Unless you know the law of sines.) You can even cross it out , if you want.

Step 1

Determine the appropriate form of formula

Since we want to find side c we need
$${\color{red}{a}}^2 = b^2 + c^2 -2bc \cdot cos( A )$$

Step 2

Substitute the values into the formula

$${\color{red}{a}}^2 = b^2 + c^2 -2bc \cdot cos( A ) \\ {\color{red}{a}}^2 = 9^2 + 7^2 -2 \cdot 9 \cdot 7 \cdot cos( 115 ) \\$$

Step 3

Solve

$${\color{red}{a}}^2 = 9^2 + 7^2 -2 \cdot 9 \cdot 7 \cdot cos( 115 ) \\ {\color{red}{a}}^2 = 171.05203545771545 \\ {\color{red}{a}} = \sqrt{ 171.05203545771545 } \\ {\color{red}{a}} = 13.07868630473701$$

### Word Problems

##### Problem 4

For word problems, the best thing to do , first and foremost, is draw a diagram.

Step 1

Draw picture

There are actually many ways to draw this triangle, but one version of the diagram is:

Regardless of how you draw it, the $$110^ \circ$$ degree angle should be in between the two known sides

Step 2

Set up formula and substitute the values

$$a^2 = b^2 + c^2 -2bc \cdot cos( A ) \\ {\color{red}{d}}^2 = e^2 + f^2 -2ef \cdot cos( D ) \\ {\color{red}{d}}^2 = 10^2 + 14^2 -2 \cdot 10 \cdot 14 \cdot cos( 110 ) \\$$

Step 3

Solve and round

$${\color{red}{d}}^2 = 200.23435986881276 \\ {\color{red}{d}} = \sqrt{200.23435986881276} \\ {\color{red}{d}} = 14.15041907043084 \\ {\color{red}{d}} = 14.2$$

##### Problem 5

For word problems, the best thing to do , first and foremost, is draw a diagram.

Step 1

Draw picture

There are actually many ways to draw this triangle, but one version of the diagram is:

Step 2

Set up formula and substitute the values

$$a^2 = b^2 + c^2 -2bc \cdot cos( A ) \\ z ^2 = x^2 + y^2 -2xy \cdot cos( {\color{red}{ Z}} ) \\ 21 ^2 = 30^2 + 15^2 -2 \cdot 30 \cdot 15 \cdot cos( {\color{red}{ Z }} )$$

Step 3

Solve and round

$$441 = 1125 - 900 \cdot cos( {\color{red}{ Z }} ) \\ \frac{441 - 1125}{ - 900 } = cos( {\color{red}{ Z }} ) \\ 0.76 = cos( {\color{red}{ Z }} ) \\ {\color{red}{ Z }} = cos^{-1}(.76) \\ {\color{red}{Z}} = 40.535802111316556 \\ {\color{red}{Z}} = 40.54$$

##### Problem 6

For word problems, the best thing to do, first and foremost, is draw a diagram.

Step 1

Draw picture

There are actually many ways to draw this triangle, but one version of the diagram is: Note, you only need one of the two angles. Can you tell which one? Click to cross out the unneeded angle

Step 2

Set up formula and substitute the values

$$a^2 = b^2 + c^2 -2bc \cdot cos( A ) \\ {\color{red}{f}} ^2 = d^2 + e^2 -2de \cdot cos( F ) \\ {\color{red}{f}} ^2 = 4^2 + 5^2 -2\cdot 4 \cdot 5 \cdot cos( 121 ^\circ )$$

Step 3

Solve and round

$${\color{red}{f}} ^2 = 4^2 + 5^2 -2\cdot 4 \cdot 5 \cdot cos( 121 ^\circ ) \\ {\color{red}{f}} ^2 = 61.60152299640217 \\ {\color{red}{f}} = \sqrt{61.60152299640217} \\ {\color{red}{f}} = 7.8486637713946035 \\ {\color{red}{f}} = 7.849$$