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SAT II Math II : Finding Angles with Trigonometry

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Example Questions

Example Question #1 : Finding Angles With Trigonometry

In $\bigtriangleup ABC$ :

AB = 23

AC = 34

BC = 25

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

Possible Answers:

$27^{\circ }$

$90^{\circ }$

$47^{\circ }$

$43^{\circ }$

$63^{\circ }$

Correct answer:

$47^{\circ }$

Explanation:

The figure referenced is below:

Triangle z

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

c = BC = 25 , the length of the side opposite the unknown angle;

a = AB = 23 ;

b = AC = 34 ;

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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SAT II Math II : Finding Angles with Trigonometry

Study concepts, example questions & explanations for SAT II Math II

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Example Question #1 : Finding Angles With Trigonometry

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