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)

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