Solve the triangles with the given parts using Law of Cosine:

Question

MN≈

m∠PNM≈ °, m∠ PM N≈ °

Answer 1

The Law of Cosines says that c²=a²+b²-2ab cos(C).

So for this problem, we first solve for MN by substituting as follows:

(MN)²=25²+38.2²-2⋅25⋅38.2⋅cos(23°)

MN=√(25²+38.2²-2⋅25⋅38.2⋅cos(23°))

MN=18.06

Always a good idea to avoid rounding errors by putting the whole thing in your calculator at once. The second line above can be put directly into most scientific or graphing calculators (make sure you are in DEG mode, not RAD) to get the result on the bottom line. If instead, you round cos(23°) and then round each subsequent step, you might end up with an answer more like MN=16.434, which is fairly far from the accurate answer.

Next, you can substitute back into the law of cosines to get other angle measures. For example,

25²=18.06²+38.2²-2⋅18.06⋅38.2⋅cos(m∠PNM)

2⋅18.06⋅38.2⋅cos(m∠PNM)=18.06²+38.2²-25²

cos(m∠PNM)=(18.06²+38.2²-25²)/(2⋅18.06⋅38.2)

m∠PNM=arccos[(18.06²+38.2²-25²)/(2⋅18.06⋅38.2)]

m∠PNM=32.75°

(Most calculators show arccos as cos^-1. I prefer to use arccos to avoid any confusion between function inverse and multiplicative inverse, which frequently use the same notation.)

Finally, you can use the fact that the interior angles of a triangle always sum to 180° to get

m∠PMN=180°-23°-32.75°

m∠PMN=124.25°