Finding Sides with Trigonometry

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SAT Subject Test in Math I › Finding Sides with Trigonometry

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In $\bigtriangleup ABC$ :

$m \angle A = 76 ^{\circ }$

$m \angle B = 65 ^{\circ }$

$\textup{Evaluate}$ $\textup{to the nearest whole unit.}$

CORRECT

Explanation

The figure referenced is below:

Triangle z

The Law of Sines states that given two angles of a triangle with measures $\alpha, \beta$ , and their opposite sides of lengths , respectively,

$\frac{\sin \beta}{b} = \frac{\sin \gamma}{c}$ ,

or, equivalently,

$\frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$ .

In this formula, we set:

, the desired sidelength;

$\beta =m \angle B = 65^{\circ }$ , the measure of its opposite angle;

, the known sidelength;

$\gamma =m \angle C$ , the measure of its opposite angle, which is

$m \angle C = 180 ^{\circ } - (m \angle A + m \angle B)$

$= 180 ^{\circ } - ( 76^{\circ } + 65 ^{\circ })$

$= 180 ^{\circ } - 141 ^{\circ }$

$=39 ^{\circ }$

Substituting in the Law of Sines formula and solving for :

$\frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$

$\frac{b}{\sin 65 ^{\circ }} = \frac{ 34}{\sin 39 ^{\circ }}$

$\frac{b}{\sin 65 ^{\circ }} \cdot \sin 65 ^{\circ } = \frac{ 34}{\sin 39 ^{\circ }} \cdot \sin 65 ^{\circ }$

$b= \frac{ 34}{\sin 39 ^{\circ }} \cdot \sin 65 ^{\circ }$

Evaluating the sines, then calculating:

$b\approx \frac{ 34}{0.6293 } \cdot 0.9063$

$b \approx 49$

A plane flies degrees north of east for miles. It then turns and flies degrees south of east for miles. Approximately how many miles is the plane from its starting point? (Ignore the curvature of the Earth.)

CORRECT

Explanation

The plane flies two sides of a triangle. The angle formed between the two sides is 40 degrees. In a Side-Angle-Side situation, it is appropriate to employ the use of the Law of Cosines.

$c^2 = a^2 + b^2 - 2ab \cos C$

$c^2 = 100^2 + 200^2 - 2(100)(200) \cos 40^{\circ}$

$c^2 \approx 19358$

$c \approx \sqrt{19358} \approx 139$

In $\bigtriangleup ABC$ :

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

$112 ^{\circ }$

CORRECT

$68^{\circ }$

$151^{\circ }$

$141^{\circ }$

Insufficient information is provided to answer the question.

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 sidelengths 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

$25^{2} =13^{2} +17^{2} - 2 (13) (17)\cos A$

$625 =169 +289- 442 \cos A$

$625 =458- 442 \cos A$

Solving for $\cos A$ :

$625 - 458 =458- 442 \cos A - 458$

$167 = - 442 \cos A$

$\frac{167}{-442} = \frac{- 442 \cos A}{-442}$

$\cos A \approx -0.3778$

Taking the inverse cosine:

$A = \cos ^{-1} -0.3778 \approx 112 ^{\circ }$ ,

the correct response.

In $\bigtriangleup ABC$ :

$m \angle A = 57^{\circ }$

Evaluate the length of $\overline{BC}$ to the nearest tenth of a unit.

CORRECT

Explanation

The figure referenced is below:

Triangle z

By the Law of Cosines, given the lengths and of two sides of a triangle, and the measure $\gamma$ of their included angle, the length of the third side can be calculated using the formula