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SAT II Math I : Trigonometry

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Example Questions

Example Question #1 : Trigonometry

A plane flies degrees north of east for 100 miles. It then turns and flies degrees south of east for 200 miles. Approximately how many miles is the plane from its starting point? (Ignore the curvature of the Earth.)

Possible Answers:

121

139

205

172

297

Correct answer:

139

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.

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

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

$c^2 \approx 19358$

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

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Example Question #1 : Trigonometry

In $\bigtriangleup ABC$ :

AB = 13

AC = 17

BC = 25

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

Possible Answers:

Insufficient information is provided to answer the question.

$151^{\circ }$

$68^{\circ }$

$112 ^{\circ }$

$141^{\circ }$

Correct answer:

$112 ^{\circ }$

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 sidelengths are known, so we are solving for $\gamma$ . Setting

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

a = AB = 13 ;

b = AC = 17 ;

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.

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Example Question #1 : Finding Sides With Trigonometry

In $\bigtriangleup ABC$ :

AB = 19

AC = 25

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

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

Possible Answers:

38.8

13.7

27.0

24.2

21.6

Correct answer:

21.6

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

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

Substituting a = AB = 19 , b = AC = 25 , c = BC , and $\gamma = m \angle A = 57^{\circ }$ , then evaluating:

$(BC)^{2} = 19 ^{2} + 25 ^{2} - 2 (19)(25)\cos 57^{\circ }$

$(BC)^{2} \approx 361 + 625 - 950 \cdot 0.5446$

$(BC)^{2} \approx 361 + 625 - 517.4$

$(BC)^{2} \approx 468.6$

Taking the square root of both sides:

$BC \approx \sqrt{468.6} \approx 21.6$ .

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Example Question #3 : Finding Sides With Trigonometry

In $\bigtriangleup ABC$ :

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

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

AB = 34

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

Possible Answers:

Correct answer:

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 a,b , respectively,

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

or, equivalently,

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

In this formula, we set:

b = AC , the desired sidelength;

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

c =AB = 34 , 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:

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

$b \approx 49$

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Example Question #1 : Trigonometry

What is the measure of the angle made between a line segment with points (0,0) , (5,8) and the -axis? Round your answer to the nearest hundreth of a degree.

Possible Answers:

$58.00^{\circ}$

No angle measure can be calculated

$71.42{^{\circ}}$

$82.87^{\circ}$

$12.42^{\circ}$

Correct answer:

$58.00^{\circ}$

Explanation:

Based on the information given, we know that the ratio of to on this segment could be represented as:

$\frac{8}{5}$

This ratio represents the tangent of the triangle formed by our line segment and the -axis. Using the inverse tangent function, we can find the angle measure:

$\tan^{-1}(\frac{8}{5}) = 57.9946167919165$

This refers to a reference angle of $58.00^{\circ}$

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Example Question #2 : Trigonometry

What is the measure of the angle made between a line segment with points (0,0) , (-8,7) and the -axis? Round your answer to the nearest hundreth of a degree.

Possible Answers:

$71.41^{\circ}$

$-48.81^{\circ}$

$41.10^{\circ}$

$48.81^{\circ}$

No angle can be calculated

Correct answer:

$48.81^{\circ}$

Explanation:

Based on the information given, we know that the ratio of to on this segment could be represented as:

$\frac{-8}{7}$

This ratio represents the tangent of the triangle formed by our line segment and the -axis. Using the inverse tangent function, we can find the angle measure:

$\tan^{-1}(\frac{-8}{7}) = -48.81407483429027$

This refers to a reference angle of $48.81^{\circ}$ .

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Example Question #2 : Trigonometry

A triangle is formed by connecting the points $(0,0), (1,0), \textup{ and }(1,6)$ . Determine the elevation angle to the nearest integer in degrees.

Possible Answers:

Correct answer:

Explanation:

After connecting the points on the graph, the length of the triangular base is 1 unit.

The height of the triangle is 6. To find the elevation angle, the angle is opposite from the height of the triangle. Since we know the base and the height, the elevation angle can be solved by using the property of tangent.

$tan(\theta)= \frac{\textup{Opposite side}}{\textup{Adjacent side}}$

$tan(\theta)= \frac{\textup{6}}{\textup{1}}$

$\theta = tan^{-1}(6)=80.53768 \textup{ degrees}$

The best answer is .

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Example Question #3 : Trigonometry

Solve for $\theta$ between $[0,2\pi]$ .

$\sin (2\theta )=\frac{1}{2}$

Possible Answers:

$\frac{\pi}{6},\frac{5\pi}{6}$

$\frac{\pi}{12},\frac{5\pi}{12}$

$\frac{\pi}{6},\frac{\pi}{3}$

$\frac{\pi}{3},\frac{2\pi}{3}$

Correct answer:

$\frac{\pi}{12},\frac{5\pi}{12}$

Explanation:

First we must solve for when sin is equal to 1/2. That is at

$\frac{\pi}{6},\frac{5\pi}{6}$

Now, plug it in:

$2\theta =\frac{\pi}{6}\Rightarrow \theta =\frac{\pi}{12}$

$2\theta =\frac{5\pi}{6}\Rightarrow \theta =\frac{5\pi}{12}$

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Example Question #4 : Trigonometry

Solve for $\theta$ between $[0,2\pi]$ .

$\cos (2\theta )=\frac{1}{2}$

Possible Answers:

$\frac{\pi}{3},\frac{2\pi}{3}$

$\frac{\pi}{6},\frac{5\pi}{6}$

$\frac{\pi}{3},\frac{5\pi}{3}$

$\frac{\pi}{12},\frac{5\pi}{12}$

Correct answer:

$\frac{\pi}{6},\frac{5\pi}{6}$

Explanation:

First we must solve for when sin is equal to 1/2. That is at

$\frac{\pi}{3},\frac{5\pi}{3}$

Now, plug it in:

$2\theta =\frac{\pi}{3}\Rightarrow \theta =\frac{\pi}{6}\,$

$2\theta =\frac{5\pi}{3}\Rightarrow \theta =\frac{5\pi}{6}$

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Example Question #7 : Trigonometry

In a triangle, ABC , what is the measure of angle A if the side opposite of $\angleA$ angle A is 3 and the adjacent side to angle A is 4?

(Round answer to the nearest tenth of a degree.)

Possible Answers:

$34.7^{\circ}$

$36.9^{\circ}$

$40^{\circ}$

$32^{\circ}$

Correct answer:

$36.9^{\circ}$

Explanation:

To find the measure of angle of A we will use tangent to solve for A. We know that

$tan A=\frac{opposite}{adjacent}$

In our case opposite = 3 and adjacent = 4, we substitute these values in and get:

$tan A= \frac{3}{4}$

Now we take the inverse tangent of each side to find the degree value of A.

$A= tan^{^{-1}}\left(\frac{3}{4}\right)$

A=36.9

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SAT II Math I : Trigonometry

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

All SAT II Math I Resources

Example Questions

Example Question #1 : Trigonometry

Example Question #1 : Trigonometry

Example Question #1 : Finding Sides With Trigonometry

Example Question #3 : Finding Sides With Trigonometry

Example Question #1 : Trigonometry

Example Question #2 : Trigonometry

Example Question #2 : Trigonometry

Example Question #3 : Trigonometry

Example Question #4 : Trigonometry

Example Question #7 : Trigonometry

All SAT II Math I Resources

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