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 34 Page 676

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

MN ≈ 48.8
m ∠ M ≈ 19
m ∠ N ≈ 28

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 later the Law of Sines.

We will first find the length of the segment MN and then the measures of the missing angles, m ∠ M and m ∠ N one at a time.

Finding MN

Note that we know an interior angle and the lengths of both of the adjacent sides. Therefore, we can use the Law of Cosines. o^2=m^2+n^2 -2 m n cos O Let's substitute m∠ O= 133, m= 21.5, and n= 31.5 to isolate o.
o^2=m^2+n^2 -2 m n cos O
o^2= 21.5^2+ 31.5^2 -2 ( 21.5)( 31.5) cos 133
Solve for o
o^2=462.25+992.25-2(21.5)(31.5)cos 133
o^2= 462.25+992.25 -1354.5cos 133
o^2=1454.5-1354.5 cos 133
o^2=1454.5-1354.5 (- 0.681998...)
o^2=1454.5-(- 923.766778...)
o^2= 2378.266779...
o=sqrt(2378.266779...)
o =48.767476 ...
o ≈ 48.8

Finding m∠ M

Now, we know the m ∠ O and the length of the side which is opposite to this angle. We can find m∠ M by using The Law of Sines! sin O/o =sin M/m Let's substitute o ≈ 48.8, m ∠ O= 133, and m= 21.5 to isolate sin M.
sin O/o =sin M/m
sin 133/48.8 = sin M/21.5
Solve for sin M
21.5 * sin 133/48.8 = sin M
sin M = 21.5 * sin 133/48.8
Now we can use the inverse sine ratio to find m ∠ M.
m ∠ M = sin ^(-1) 21.5 * sin 133/48.8
m ∠ M ≈ 18.796947
m ∠ M ≈ 19

Finding m ∠ N

Finally, to find m ∠ N we can use the Triangle Angle Sum Theorem. This tells us that the measures of the angles in a triangle add up to 180. 133+ 19+ m ∠ N= 180 ⇔ m ∠ N≈ 28

Completing the Triangle

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