Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
5. The Law of Cosines
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Exercise 3 Page 939

The Law of Cosines relates the cosine of each angle of a triangle to its side lengths.

34.1 ^(∘)

Practice makes perfect

For any △ ABC, the Law of Cosines relates the cosine of each angle to the side lengths of the triangle.

To find the missing angle measure, we will start by drawing a diagram to illustrate the situation.

We know that the lengths of PN and KP are 21 and 12, respectively. We also know that their included angle has a measure of 67^(∘). To find the measure of ∠ N we need to find KN first. Let's do this one at a time.

Finding KN

With the provided information and using the Law of Cosines, we can write an equation to find p. Then, we will be able to use the Law of Cosines once again for ∠ N. p^2 = k^2 + n^2 - 2 kn cos P Let's substitute k= 21, n = 12, and m ∠ P = 67^(∘).
p^2 = 21^2 + 12^2 - 2( 21)( 12)cos 67^(∘)
Solve for p
p^2=441 + 144 - 2(21)(12)cos 67^(∘)
p^2 = 441 + 144 - 2(21)(12)(0.390 731...)
p^2= 441 + 144 - 196.928 489...
p^2= 388.071 511...
p = sqrt(388.071 511...)
p = 19.699531...
p ≈ 19.7
Note that we only kept the principal root when solving the equation because p is the length of a side.

Finding m∠ N

Now, we are able to use the Law of Cosines once again for the angle N.
12^2 = 21^2 + 19.7^2 -2( 21)( 19.7)cos N
Solve for cos N
144 = 441 + 388.09 - 2(21)(19.7)cosN
144 = 441 + 388.09 - 827.4cosN
144 = 829.09 - 827.4cosN
- 685.09 = - 827.4 cosN
- 685.09/- 827.4= cos N
685.09/827.4 =cos N
cos N = 685.09/827.4
To find m∠ N, we will use the inverse operation of cos, which is cos ^(- 1). cos N = 685.09/827.4 ⇕ m∠ N = cos ^(- 1)(685.09/827.4) Finally, we will use a calculator.
m∠ N = cos ^(- 1)(685.09/827.4)
m∠ N = 34.105819...^(∘)
m∠ N≈ 34.1^(∘)