McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. The Law of Sines and Law of Cosines
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Exercise 37 Page 676

Begin by using the Law of Cosines first and then the Law of Sines.

PR ≈ 14.6
m ∠ P ≈ 35
m ∠ R ≈ 75

Practice makes perfect

Let's begin by color coding the opposite angles, sides, and the vertices in the given triangle. It will help us use the Law of Cosines and the Law of Sines later.

Let's first find the length of the missing segment PR and then the measures of the missing angles, m ∠ P and m ∠ R one at a time.

Finding PR

Note that we know an interior angle and the lengths of both of the adjacent sides. Therefore, we can use the Law of Cosines. q^2=p^2+r^2 -2 p r cos Q Let's substitute m∠ Q= 70, p= 9, and r= 15 to isolate q.
q^2=p^2+r^2 -2 p r cos Q
q^2= 9^2+ 15^2 -2 ( 9)( 15) cos 70
Solve for q
q^2=81+225-2(9)(15)cos 70
q^2= 81+225-270cos 70
q^2=306-270cos 70
q^2= 213.654561...
q=sqrt(213.654561...)
q =14.616927 ...
q ≈ 14.6

Finding m∠ P

Now, we know the m ∠ Q and the length of the side which is opposite to this angle. We can find the measure of m∠ P by using the Law of Sines. sin Q/q =sin P/p Let's substitute q ≈ 14.6, m ∠ Q= 70, and p= 9 to isolate sin P.
sin Q/q =sin P/p
sin 70/14.6 = sin P/9
Solve for sin P
9 sin 70/14.6 = sin P
sin P = 9 sin 70/14.6
Now we can use the inverse sine ratio to find m ∠ P.
m ∠ P = sin ^(-1) 9 sin 70/14.6
m ∠ P ≈ 35.398692
m ∠ P ≈ 35

Finding m ∠ R

Finally, to find m ∠ R we can use the Triangle Angle Sum Theorem. This tells us that the measures of the angles in a triangle add up to 180. 35 + 70 + m ∠ R = 180 ⇔ m ∠ R≈ 75

Completing the Triangle

With all of the angle measures and side lengths, we can complete our diagram.