Finding Angles with Trigonometry - SAT Subject Test in Math II

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Question

In $\bigtriangleup ABC$ :

Evaluate $m \angle A$ to the nearest degree.

Answer

The figure referenced is below:

By the Law of Cosines, the relationship of the measure of an angle $\gamma$ of a triangle and the three side lengths , , and , the sidelength opposite the aforementioned angle, is as follows:

$c^{2} = a ^{2} + b^{2} - 2ab \cos \gamma$

All three side lengths are known, so we are solving for $\gamma$ . Setting

, the length of the side opposite the unknown angle;

;

;

and $\gamma = \cos A$ ,

We get the equation

$25^{2} =23^{2} +34^{2} - 2 (23) (34)\cos A$

$625 =529+1,156- 1,564 \cos A$

$625 =1,685 - 1,564 \cos A$

Solving for $\cos A$ :

$625 - 1,685 =1,685 - 1,564 \cos A - 1,685$

$-1.060 = - 1,564 \cos A$

$\frac{-1.060 }{- 1,564}= \frac{- 1,564 \cos A}{- 1,564}$

$\cos A \approx 0.6777$

Taking the inverse cosine:

$A = \cos ^{-1} 0.6777 \approx 47 ^{\circ }$ ,

the correct response.

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