AP Precalculus -3.2 Sine, Cosine, and Tangent- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.2 Sine, Cosine, and Tangent- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.2 Sine, Cosine, and Tangent- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

In the $ xy $-plane, an angle in standard position measures $\frac{5\pi}{6}$ radians. A circle centered at the origin has radius 4. What are the coordinates of the point of intersection of the terminal ray of the angle and the circle?

(A) $(-2\sqrt{3}, 2)$
(B) $(-\frac{\sqrt{3}}{2}, \frac{1}{2})$
(C) $(\frac{\sqrt{3}}{2}, \frac{1}{2})$
(D) $(2\sqrt{3}, 2)$

▶️ Answer/Explanation

Detailed solution

For a point on a circle of radius $ r = 4 $ at angle $ \theta = \frac{5\pi}{6} $:
$ x = r \cos\theta = 4 \cos\left(\frac{5\pi}{6}\right) = 4 \cdot \left(-\frac{\sqrt{3}}{2}\right) = -2\sqrt{3} $
$ y = r \sin\theta = 4 \sin\left(\frac{5\pi}{6}\right) = 4 \cdot \frac{1}{2} = 2 $
Thus, the intersection point is $ (-2\sqrt{3}, 2) $, which is in Quadrant II.
✅ Answer: (A)

Question

The figure gives a right triangle, where $ y $ is the length of the side opposite angle $ A $. If the function $ f $ gives values of $ A $ as a function of $ y $, which of the following could define $ f(y) $?

(A) $ f(y) = \cos\left(\frac{y}{12}\right) $
(B) $ f(y) = \sin\left(\frac{y}{12}\right) $
(C) $ f(y) = \cos^{-1}\left(\frac{y}{12}\right) $
(D) $ f(y) = \sin^{-1}\left(\frac{y}{12}\right) $

▶️ Answer/Explanation

Detailed solution

In a right triangle, the sine of angle $ A $ is the ratio of the opposite side $ y $ to the hypotenuse. From the figure, the hypotenuse is 12, so $ \sin A = \frac{y}{12} $.
To express $ A $ as a function of $ y $, we take the inverse sine: $ A = \sin^{-1}\left(\frac{y}{12}\right) $.
Thus, $ f(y) = \sin^{-1}\left(\frac{y}{12}\right) $.
✅ Answer: (D)

Question

The figure shows a circle of radius 2 along with four labeled points in the $xy$-plane. The measure of angle $COB$ is equal to the measure of angle $AOB$. What are the coordinates of point $B$?

(A) $\left(\cos\left(\frac{7\pi}{4}\right), \sin\left(\frac{7\pi}{4}\right)\right)$
(B) $\left(\sin\left(\frac{7\pi}{4}\right), \cos\left(\frac{7\pi}{4}\right)\right)$
(C) $\left(2\cos\left(\frac{7\pi}{4}\right), 2\sin\left(\frac{7\pi}{4}\right)\right)$
(D) $\left(2\sin\left(\frac{7\pi}{4}\right), 2\cos\left(\frac{7\pi}{4}\right)\right)$

▶️ Answer/Explanation

Detailed solution

From the figure (assuming standard positions):
– Angle $ AOB $ is measured from positive x-axis to ray $ OB $.
– Given $ \angle COB = \angle AOB $, and likely $ A $ is on positive x-axis, $ C $ is on positive y-axis or similar.
– If $ OA $ is along positive x-axis, $ OC $ along positive y-axis, then $ \angle AOB = \angle COB $ implies $ OB $ bisects the quadrant.
For Quadrant IV: $ OB $ at $ \frac{7\pi}{4} $ (if starting from +x, rotating clockwise to match symmetry).
Radius is 2, so coordinates: $ (2\cos\frac{7\pi}{4}, 2\sin\frac{7\pi}{4}) $.
✅ Answer: (C)

Question

Angles $ A $ and $ B $ are in standard position in the $ xy $-plane. The measure of angle $ A $ is $ \frac{2\pi}{3} $ radians, and the measure of angle $ B $ is $ \frac{4\pi}{3} $ radians. The terminal rays of both angles intersect a circle centered at the origin with radius 20. What is the distance between these two points of intersection?

(A) $ \cos\left(\frac{2\pi}{3}\right) – \cos\left(\frac{4\pi}{3}\right) $
(B) $ 20\cos\left(\frac{2\pi}{3}\right) – 20\cos\left(\frac{4\pi}{3}\right) $
(C) $ \sin\left(\frac{2\pi}{3}\right) – \sin\left(\frac{4\pi}{3}\right) $
(D) $ 20\sin\left(\frac{2\pi}{3}\right) – 20\sin\left(\frac{4\pi}{3}\right) $

▶️ Answer/Explanation

Detailed solution

Points:
$ A: (20\cos\frac{2\pi}{3}, 20\sin\frac{2\pi}{3}) = \left(20\cdot\left(-\frac12\right), 20\cdot\frac{\sqrt{3}}{2}\right) = (-10, 10\sqrt{3}) $
$ B: (20\cos\frac{4\pi}{3}, 20\sin\frac{4\pi}{3}) = \left(20\cdot\left(-\frac12\right), 20\cdot\left(-\frac{\sqrt{3}}{2}\right)\right) = (-10, -10\sqrt{3}) $
The points are vertically aligned (same $ x $-coordinate).
Distance = $ |10\sqrt{3} – (-10\sqrt{3})| = 20\sqrt{3} $.
That’s equal to $ 20\sin\frac{2\pi}{3} – 20\sin\frac{4\pi}{3} $ because:
$ 20\sin\frac{2\pi}{3} = 20\cdot\frac{\sqrt{3}}{2} = 10\sqrt{3} $
$ 20\sin\frac{4\pi}{3} = 20\cdot\left(-\frac{\sqrt{3}}{2}\right) = -10\sqrt{3} $
Difference: $ 10\sqrt{3} – (-10\sqrt{3}) = 20\sqrt{3} $, which is the vertical distance.
✅ Answer: (D)

Question

The figure shows a spinner with eight congruent sectors in the $xy$-plane. The origin and three points are labeled. If the coordinates of $X$ are $\left(-\frac{5}{2}, -\frac{5\sqrt{3}}{2}\right)$, what is the measure of angle $XOB$?

(A) $\frac{\pi}{6}$
(B) $\frac{\pi}{3}$
(C) $\frac{7\pi}{6}$
(D) $\frac{4\pi}{3}$

▶️ Answer/Explanation

Detailed solution

Given $ X = \left(-\frac{5}{2}, -\frac{5\sqrt{3}}{2}\right) $.
Radius $ r = \sqrt{\left(\frac{5}{2}\right)^2 + \left(\frac{5\sqrt{3}}{2}\right)^2} = \sqrt{\frac{25}{4}+\frac{75}{4}} = \sqrt{25} = 5 $.
The coordinates match $ \left(5\cdot\left(-\frac12\right), 5\cdot\left(-\frac{\sqrt{3}}{2}\right)\right) $, so $ \cos\theta = -\frac12, \sin\theta = -\frac{\sqrt{3}}{2} $.
That’s $ \theta = \frac{4\pi}{3} $ (Quadrant III).
$ B $ is at angle $ \pi $ (since it’s at $(-5,0)$).
Angle $ XOB $ = $ \theta_X – \theta_B = \frac{4\pi}{3} – \pi = \frac{\pi}{3} $.
✅ Answer: (B)

Question

In the $ xy $-plane, an angle $ \theta $ in standard position has measure $ \theta = \frac{\pi}{3} $. Which of the following is true?

(A) The slope of the terminal ray is $ \frac{1}{2} $.
(B) The slope of the terminal ray is $ \frac{1}{\sqrt{3}} $.
(C) The slope of the terminal ray is $ \frac{\sqrt{3}}{2} $.
(D) The slope of the terminal ray is $ \sqrt{3} $.

▶️ Answer/Explanation

Detailed solution

The slope of the terminal ray of an angle in standard position is given by $ \tan\theta $.
For $ \theta = \frac{\pi}{3} $, $ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} $.
✅ Answer: (D)

Question

Angle $ BAC $ intersects unit circle at $ (0.4, -0.9) $. Which is correct about tangent and slope?

(A) Tangent $ -\frac{4}{9} $, slope $ -\frac{4}{9} $
(B) Tangent $ -\frac{4}{9} $, slope $ -\frac{9}{4} $
(C) Tangent $ -\frac{9}{4} $, slope $ -\frac{9}{4} $
(D) Tangent $ -\frac{9}{4} $, slope $ \frac{4}{9} $

▶️ Answer/Explanation

Detailed solution

Tangent = $ \frac{y}{x} = \frac{-0.9}{0.4} = -\frac{9}{4} $.
Slope of terminal ray = tangent = $ -\frac{9}{4} $.
✅ Answer: (C)

Question

In the $xy$-plane, two different angles $\alpha$ and $\beta$ in standard position share a terminal ray. Which of the following gives possible values for $\alpha$ and $\beta$?

(A) $\alpha=-\frac{\pi}{4}, \beta=-\frac{7\pi}{4}$
(B) $\alpha=\frac{3\pi}{5}, \beta=-\frac{3\pi}{5}$
(C) $\alpha=\frac{2\pi}{3}, \beta=\frac{8\pi}{3}$
(D) $\alpha=\frac{5\pi}{6}, \beta=-\frac{\pi}{6}$

▶️ Answer/Explanation

Detailed solution

1. Condition for Coterminal Angles:
Angles share a terminal ray if their difference is a multiple of $2\pi$.

2. Check Option (C):
$\frac{8\pi}{3} – \frac{2\pi}{3} = \frac{6\pi}{3} = 2\pi$.
Since the difference is exactly $2\pi$, they are coterminal.

✅ Answer: (C)

Question

The figure shows a circle centered at the origin with an angle of measure $\theta$ radians in standard position and point $P$ on the circle. The terminal ray of the angle intersects the circle at point $Q$. The length of arc $\overline{PQ}$ is 6 units. Which of the following gives the distance of point $Q$ from the $y$-axis?

(A) $cos(\frac{6}{5})$
(B) $sin(\frac{6}{5})$
(C) $5~cos(\frac{6}{5})$
(D) $5~sin(\frac{6}{5})$

▶️ Answer/Explanation

Detailed solution

1. Determine the Radius:
Point $P$ is at $(5,0)$ and is on the circle. Thus, the radius $r = 5$.

2. Determine the Angle $\theta$:
The arc length formula is $s = r\theta$.
Given $s = 6$ and $r = 5$, we have $6 = 5\theta$, so $\theta = \frac{6}{5}$ radians.

3. Find the Distance from the Y-axis:
The distance of a point $(x, y)$ from the y-axis is $|x|$.
The x-coordinate of point $Q$ is $x = r\cos\theta = 5\cos(\frac{6}{5})$.
Therefore, the distance is $5\cos(\frac{6}{5})$.

✅ Answer: (C)

Question

The figure shows a circle centered at the origin with an angle of measure $\theta$ radians in standard position. The terminal ray of the angle intersects the circle at point $P$. If $\tan\theta = -\frac{3}{4}$, what is the slope of the line that passes through the origin and $P$?

(A) $-\frac{4}{3}$
(B) $-\frac{4}{5}$
(C) $-\frac{3}{4}$
(D) $-\frac{3}{5}$

▶️ Answer/Explanation

Detailed solution

1. Definition:
Slope $m$ of terminal ray = $\tan\theta$.

2. Calculation:
Given $\tan\theta = -\frac{3}{4}$.
Slope $m = -\frac{3}{4}$.

✅ Answer: (C)

Question

Given the point $(\sqrt{17}, -8)$ on the terminal angle $\theta$, what is $\tan \theta$?

a. $-\frac{\sqrt{5}}{3}$
b. $-\frac{3\sqrt{5}}{5}$
c. $-\frac{9\sqrt{17}}{17}$
d. $-\frac{8\sqrt{17}}{17}$

▶️ Answer/Explanation

Detailed solution

Identify the coordinates of the given point: $x = \sqrt{17}$ and $y = -8$.
Recall the trigonometric ratio for tangent: $\tan \theta = \frac{y}{x}$.
Substitute the values: $\tan \theta = \frac{-8}{\sqrt{17}}$.
Rationalize the denominator by multiplying the top and bottom by $\sqrt{17}$.
Calculation: $\frac{-8 \cdot \sqrt{17}}{\sqrt{17} \cdot \sqrt{17}} = -\frac{8\sqrt{17}}{17}$.
Compare the result with the given choices.
The correct option is d.

Question

A circle of unknown radius has a terminal point at angle $\theta = \frac{\pi}{5}$. What is the slope of the line tangent to this terminal point on the circle?

a. $-0.73$

b. $-1.38$

c. $-1.70$

d. $-1.24$

▶️ Answer/Explanation

Detailed solution

The slope of the radial line from the origin $(0,0)$ to the terminal point is $m_{radius} = \tan(\theta)$.
Substituting the given angle, the radial slope is $m_{radius} = \tan\left(\frac{\pi}{5}\right) \approx 0.7265$.
A tangent line to a circle is perpendicular to the radius at the point of contact.
Perpendicular slopes are negative reciprocals: $m_{tangent} = -\frac{1}{m_{radius}}$.
Thus, $m_{tangent} = -\frac{1}{\tan(\pi/5)} = -\cot\left(\frac{\pi}{5}\right)$.
Calculating the value: $-\frac{1}{0.7265} \approx -1.376$.
Rounding to two decimal places, the slope is $-1.38$.
The correct option is b.

AP Precalculus -3.2 Sine, Cosine, and Tangent- MCQ Exam Style Questions - Effective Fall 2023