Trigonometry Practice Questions — ACT — Page 3

Question 41Question →

A surveyor is measuring a triangular plot of land, $ABC$ . The distance from point $A$ to point $C$ is $12$ meters, and the distance from point $B$ to point $C$ is $6\sqrt{2}$ meters. If the measure of angle $BAC$ is $30^\circ$ , what is the measure, in degrees, of the acute angle $ABC$ ?

Show answer & explanation

Answer: 45

Answer

The measure of the acute angle

ABC

is

45

degrees.

Applying the Law of Sines yields the relation

\frac{\sin(B)}{12} = \frac{\sin(30^\circ)}{6\sqrt{2}}

. Solving for

\sin(B)

yields

\sin(B) = \frac{\sqrt{2}}{2}

. Because the question specifies that the angle is acute, the measure of the angle is

45^\circ

.

Step-by-Step Solution

1

Set up the Law of Sines relationship for triangle

ABC

using the side opposite angle

B

(

AC

) and the side opposite angle

A

(

BC

).

\frac{\sin(B)}{AC} = \frac{\sin(A)}{BC}

The Law of Sines states that the ratio of the sine of an angle to the length of its opposite side is constant for all three angles in a triangle.

2

Substitute the given values into the equation:

AC = 12

,

BC = 6\sqrt{2}

, and

A = 30^\circ

.

\frac{\sin(B)}{12} = \frac{\sin(30^\circ)}{6\sqrt{2}}

This allows us to solve for the single unknown variable, the sine of angle

B

.

3

Simplify the expression using the trigonometric value

\sin(30^\circ) = 0.5

and isolate

\sin(B)

.

\sin(B) = \frac{12 \cdot 0.5}{6\sqrt{2}} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Simplifying the fractions helps us identify the standard trigonometric value.

4

Find the acute angle

B

whose sine value is

\frac{\sqrt{2}}{2}

.

B = 45^\circ

The inverse sine of

\frac{\sqrt{2}}{2}

for an acute angle is

45^\circ

.

Key Concept

Law of Sines

Estimated Time:1m 0s

Question 42Question →

An angle $\theta$ is positioned in the second quadrant. If the trigonometric expression $\csc\theta - \cot\theta$ is equal to $4$ , what is the value of $17\cos\theta$ ?

Show answer & explanation

Answer: -15

Answer

The value of

17\cos\theta

is

-15

.

The correct value is

-15

. By using the difference of squares on the identity

\csc^2\theta - \cot^2\theta = 1

, we establish that

\csc\theta + \cot\theta = \frac{1}{4}

. Solving this system alongside

\csc\theta - \cot\theta = 4

yields

\csc\theta = \frac{17}{8}

and

\cot\theta = -\frac{15}{8}

. Since

\cos\theta = \frac{\cot\theta}{\csc\theta}

, we find

\cos\theta = -\frac{15}{17}

, which means

17\cos\theta = -15

.

Step-by-Step Solution

1

Use the Pythagorean identity linking cosecant and cotangent to find the sum of the terms.

\csc\theta + \cot\theta = \frac{1}{4}

Since

\csc^2\theta - \cot^2\theta = 1

, factoring as a difference of squares gives

(\csc\theta - \cot\theta)(\csc\theta + \cot\theta) = 1

. Substituting

\csc\theta - \cot\theta = 4

yields

4(\csc\theta + \cot\theta) = 1

, so

\csc\theta + \cot\theta = \frac{1}{4}

.

2

Solve the system of equations for

\csc\theta

and

\cot\theta

.

\csc\theta = \frac{17}{8}

and

\cot\theta = -\frac{15}{8}

Adding the equations

\csc\theta - \cot\theta = 4

and

\csc\theta + \cot\theta = \frac{1}{4}

gives

2\csc\theta = \frac{17}{4} \implies \csc\theta = \frac{17}{8}

. Subtracting the first equation from the second gives

2\cot\theta = -\frac{15}{4} \implies \cot\theta = -\frac{15}{8}

.

3

Determine the value of

\cos\theta

using reciprocal and quotient identities.

\cos\theta = -\frac{15}{17}

Using the relationship

\cos\theta = \frac{\cot\theta}{\csc\theta}

, we substitute the values to find

\cos\theta = \frac{-15/8}{17/8} = -\frac{15}{17}

.

4

Calculate the final required expression value.

17\cos\theta = -15

Multiplying the calculated value of

\cos\theta

by

17

gives

17 \left(-\frac{15}{17}\right) = -15

.

Key Concept

Using fundamental Pythagorean, reciprocal, and quotient trigonometric identities to solve systems of equations and evaluate trigonometric expressions.

Question 43Question →

In right triangle $ABC$ , the right angle is at vertex $C$ . The length of leg $AC$ is $12$ inches. If $\sin(B) = \frac{3}{5}$ , what is the length, in inches, of the hypotenuse $AB$ ?

9

15

16

20

25

Show answer & explanation

Answer: 20

Answer

The length of the hypotenuse

AB

is 20 inches.

The sine of angle

B

is defined as the ratio of the opposite side to the hypotenuse, which is

\sin(B) = \frac{AC}{AB}

. Substituting the given values, we get

\frac{3}{5} = \frac{12}{AB}

. Solving for the hypotenuse

AB

gives

3 \cdot AB = 60

, which simplifies to

AB = 20

.

Step-by-Step Solution

1

Identify the trigonometric ratio for sine in a right triangle.

\sin(B) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{AB}

By definition of right triangle trigonometry (SOHCAHTOA), the sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse.

2

Substitute the given values into the sine ratio equation.

\frac{3}{5} = \frac{12}{AB}

The opposite side to angle

B

is leg

AC

, which has a length of

12

inches, and

\sin(B)

is given as

\frac{3}{5}

.

3

Solve the proportion for the hypotenuse

AB

.

3 \cdot AB = 60 \implies AB = 20

Cross-multiplying yields

3 \cdot AB = 5 \cdot 12 = 60

. Dividing both sides by

3

gives the length of the hypotenuse.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:45s

Question 44Question →

For an angle $\theta$ satisfying $\frac{\pi}{2} < \theta < \pi$ , if $\sec \theta = -\frac{13}{5}$ , what is the value of $12(\cot \theta + \csc \theta)$ ?

Show answer & explanation

Answer: 8

Answer

8

For an angle

\theta

in Quadrant II (

\frac{\pi}{2} < \theta < \pi

), cosine is negative and sine is positive. Given

\sec \theta = -\frac{13}{5}

, the reciprocal identity gives

\cos \theta = -\frac{5}{13}

. The Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

yields

\sin \theta = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \frac{12}{13}

. Then

\cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{5}{12}

and

\csc \theta = \frac{1}{\sin \theta} = \frac{13}{12}

. Adding these values gives

\cot \theta + \csc \theta = \frac{8}{12} = \frac{2}{3}

. Multiplying by 12 yields the final value of 8.

Step-by-Step Solution

1

Find

\cos \theta

from

\sec \theta

\cos \theta = -\frac{5}{13}

By definition of the reciprocal trigonometric identity,

\cos \theta = \frac{1}{\sec \theta}

.

2

Calculate

\sin \theta

using the Pythagorean identity

\sin \theta = \frac{12}{13}

In Quadrant II (

\frac{\pi}{2} < \theta < \pi

), sine is positive. Applying

\sin \theta = \sqrt{1 - \cos^2 \theta}

gives

\sqrt{1 - \frac{25}{169}} = \frac{12}{13}

.

3

Find

\cot \theta

and

\csc \theta

\cot \theta = -\frac{5}{12} \text{ and } \csc \theta = \frac{13}{12}

Using quotient identity

\cot \theta = \frac{\cos \theta}{\sin \theta}

and reciprocal identity

\csc \theta = \frac{1}{\sin \theta}

.

4

Substitute into the given expression

12(\cot \theta + \csc \theta)

and simplify

8

12\left(-\frac{5}{12} + \frac{13}{12}\right) = 12\left(\frac{8}{12}\right) = 8

.

Key Concept

Pythagorean, quotient, and reciprocal identities with quadrant sign analysis

Question 45Question →

A radar signal sweeps counterclockwise around a control tower located at the origin of a coordinate plane. Starting from standard position along the positive $x$ -axis, the radar line rotates through an angle of $\frac{13\pi}{4}$ radians. Which of the following ordered pairs represents the $(x, y)$ coordinates of the point where the radar line intersects the unit circle centered at the origin?

\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)

\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

\left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)

\left(-\frac{\sqrt{3}}{2}, -\frac{1}{2}\right)

Show answer & explanation

Answer: $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$

Answer

\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

Subtracting

2\pi

(one full rotation) from

\frac{13\pi}{4}

gives the coterminal angle

\frac{5\pi}{4}

. This angle terminates in Quadrant III, where both cosine (

x

-coordinate) and sine (

y

-coordinate) are negative. Using the reference angle

\frac{\pi}{4}

, both values have magnitude

\frac{\sqrt{2}}{2}

, giving the coordinate pair

\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

.

Step-by-Step Solution

1

Find an equivalent coterminal angle within one full revolution

[0, 2\pi)

.

Subtract

2\pi = \frac{8\pi}{4}

from

\frac{13\pi}{4}

:

\frac{13\pi}{4} - \frac{8\pi}{4} = \frac{5\pi}{4}

radians.

Coterminal angles share the exact same terminal ray and unit circle coordinates.

2

Identify the quadrant and reference angle for

\frac{5\pi}{4}

.

The angle lies in Quadrant III because

\pi < \frac{5\pi}{4} < \frac{3\pi}{2}

. The reference angle is

\frac{5\pi}{4} - \pi = \frac{\pi}{4}

.

The reference angle determines the absolute magnitude of the trigonometric coordinates.

3

Calculate the coordinates

(x, y) = (\cos\theta, \sin\theta)

for the terminal ray.

Since

\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

and

\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}

, and both coordinates are negative in Quadrant III,

(x, y) = \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)

.

Points on the unit circle are defined by

(\cos\theta, \sin\theta)

.

Key Concept

Coterminal Angles and Unit Circle Coordinates

Estimated Time:1m 0s

Question 46Question →

In triangle $XYZ$ , the length of side $XY$ is $14$ meters, the measure of $\angle X$ is $40^\circ$ , and the measure of $\angle Z$ is $80^\circ$ . Which of the following expressions represents the length, in meters, of side $YZ$ ?

\frac{14 \sin(40^\circ)}{\sin(80^\circ)}

\frac{14 \sin(80^\circ)}{\sin(40^\circ)}

\frac{\sin(40^\circ)}{14 \sin(80^\circ)}

14 \sin(40^\circ) \sin(80^\circ)

\frac{14}{\sin(40^\circ) \sin(80^\circ)}

Show answer & explanation

Answer: $\frac{14 \sin(40^\circ)}{\sin(80^\circ)}$

Answer

\frac{14 \sin(40^\circ)}{\sin(80^\circ)}

The expression derived by applying the Law of Sines

\frac{a}{\sin(A)} = \frac{c}{\sin(C)}

correctly matches side

YZ

with its opposite angle

\angle X = 40^\circ

and side

XY = 14

with its opposite angle

\angle Z = 80^\circ

, yielding

YZ = \frac{14 \sin(40^\circ)}{\sin(80^\circ)}

.

Step-by-Step Solution

1

Identify the relevant law and relate the known sides and angles

Using the Law of Sines:

\frac{YZ}{\sin(X)} = \frac{XY}{\sin(Z)}

The Law of Sines relates the side lengths of a triangle to the sines of their opposite angles.

2

Substitute the given values into the formula

\frac{YZ}{\sin(40^\circ)} = \frac{14}{\sin(80^\circ)}

Side

XY = 14

is opposite

\angle Z = 80^\circ

, and side

YZ

is opposite

\angle X = 40^\circ

.

3

Solve for the unknown side

YZ

YZ = \frac{14 \sin(40^\circ)}{\sin(80^\circ)}

Multiply both sides of the equation by

\sin(40^\circ)

to isolate

YZ

.

Key Concept

Law of Sines

Question 47Question →

The graph of the function $f(x) = A \sin(Bx + C) + D$ has a minimum point at $\left(\frac{\pi}{4}, -2\right)$ and its consecutive maximum point at $\left(\frac{5\pi}{4}, 8\right)$ , where $A > 0$ and $B > 0$ . What is the value of $f\left(\frac{7\pi}{4}\right)$ ?

-2

0

3

5

8

Show answer & explanation

Answer: 3

Answer

3

The midline of a sinusoidal curve is the average of its maximum and minimum values, which is

\frac{8 + (-2)}{2} = 3

. The horizontal distance between consecutive minimum and maximum points is half of a period:

\frac{5\pi}{4} - \frac{\pi}{4} = \pi

. Thus, the full period is

2\pi

, and the next minimum occurs at

\frac{5\pi}{4} + \pi = \frac{9\pi}{4}

. The value

x = \frac{7\pi}{4}

lies halfway between the maximum at

\frac{5\pi}{4}

and the minimum at

\frac{9\pi}{4}

. Midway between a maximum and a minimum, the function crosses its midline, so

f\left(\frac{7\pi}{4}\right) = 3

.

Step-by-Step Solution

1

Determine the midline (vertical shift

D

) of the trigonometric function.

The midline is

D = \frac{\text{Maximum} + \text{Minimum}}{2} = \frac{8 + (-2)}{2} = 3

.

The midline lies exactly halfway between the vertical peak and trough.

2

Find the horizontal distance for a half-period and determine the full period.

Half-period =

\frac{5\pi}{4} - \frac{\pi}{4} = \pi

, so the full period

T = 2\pi

.

The horizontal distance between consecutive minimum and maximum points equals half of one full period.

3

Identify the behavior of the graph at

x = \frac{7\pi}{4}

.

Since

x = \frac{7\pi}{4}

is halfway between the maximum at

x = \frac{5\pi}{4}

and the next minimum at

x = \frac{9\pi}{4}

, the function value equals the midline height

D = 3

.

A sinusoidal wave crosses its midline exactly midway between a peak and a trough.

Key Concept

Key features of transformed trigonometric functions (amplitude, midline, period, and symmetry)

Estimated Time:2m 0s

Question 48Question →

What is the exact value of $\cos\left(\arcsin\left(\frac{3}{5}\right)\right)$ expressed as a decimal?

Show answer & explanation

Answer: 0.8

Answer

The exact value of the expression is 0.8.

Evaluating

\cos\left(\arcsin\left(\frac{3}{5}\right)\right)

requires finding the cosine of an angle

\theta

whose sine is

\frac{3}{5}

. In a right triangle with an opposite side of 3 and a hypotenuse of 5, the adjacent side is

\sqrt{5^2 - 3^2} = 4

. The cosine of

\theta

is the ratio of the adjacent side to the hypotenuse, which gives

\frac{4}{5} = 0.8

.

Step-by-Step Solution

1

Interpret the inverse sine function as an angle.

Let

\theta = \arcsin\left(\frac{3}{5}\right)

, meaning

\sin(\theta) = \frac{3}{5}

for

0 < \theta < \frac{\pi}{2}

.

The inverse sine function returns an angle whose sine is the given value within the principal interval

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

.

2

Find the adjacent side of the right triangle associated with angle

\theta

.

\text{adjacent} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4

.

By the Pythagorean theorem,

a^2 + b^2 = c^2

, so the adjacent side length is

\sqrt{c^2 - b^2}

.

3

Calculate the cosine of angle

\theta

.

\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5} = 0.8

.

Cosine is defined as the ratio of the adjacent side to the hypotenuse in a right triangle.

Key Concept

Composition of Trigonometric and Inverse Trigonometric Functions

Estimated Time:45s

Question 49Question →

If $\theta = \arcsin\left(\frac{3}{5}\right)$ , where $0 \leq \theta \leq \frac{\pi}{2}$ , what is the value of $\tan(\theta)$ ?

\frac{3}{4}

\frac{4}{3}

\frac{4}{5}

\frac{5}{3}

\frac{5}{4}

Show answer & explanation

Answer: $\frac{3}{4}$

Answer

\frac{3}{4}

Since

\theta = \arcsin\left(\frac{3}{5}\right)

,

\sin(\theta) = \frac{3}{5}

. In a right-angled triangle with acute angle

\theta

, the opposite side is 3 and the hypotenuse is 5. By the Pythagorean theorem, the adjacent side is

\sqrt{5^2 - 3^2} = 4

. The tangent of

\theta

is defined as the ratio of the opposite side to the adjacent side, which gives

\frac{3}{4}

.

Step-by-Step Solution

1

Interpret the inverse trigonometric equation

\sin(\theta) = \frac{3}{5}

By definition of inverse sine,

\theta = \arcsin\left(\frac{3}{5}\right)

means

\sin(\theta) = \frac{3}{5}

for

0 \leq \theta \leq \frac{\pi}{2}

.

2

Determine the side lengths of the reference right triangle

\text{opposite} = 3, \quad \text{hypotenuse} = 5, \quad \text{adjacent} = \sqrt{5^2 - 3^2} = 4

Using the ratio definition

\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}

and the Pythagorean theorem

a^2 + b^2 = c^2

to solve for the adjacent side.

3

Evaluate

\tan(\theta)

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}

The tangent function is defined as the ratio of the length of the opposite side to the length of the adjacent side.

Key Concept

Evaluating composite trigonometric expressions involving inverse functions using right triangle geometry.

Question 50Question →

If $\theta$ is an angle in Quadrant IV such that $\cos\theta = \frac{4}{5}$ , what is the value of $\frac{\tan\theta + \sec\theta}{\csc\theta}$ ?

-\frac{3}{10}

\frac{6}{5}

-\frac{6}{5}

-\frac{5}{6}

-\frac{3}{8}

Show answer & explanation

Answer: $-\frac{3}{10}$

Answer

-\frac{3}{10}

Using the Pythagorean identity

\sin^2\theta + \cos^2\theta = 1

in Quadrant IV yields

\sin\theta = -\frac{3}{5}

. Substituting this into quotient and reciprocal identities gives

\tan\theta = -\frac{3}{4}

,

\sec\theta = \frac{5}{4}

, and

\csc\theta = -\frac{5}{3}

. Evaluating

\frac{\tan\theta + \sec\theta}{\csc\theta}

gives

\frac{1/2}{-5/3} = -\frac{3}{10}

.

Step-by-Step Solution

1

Determine

\sin\theta

using the Pythagorean identity

\sin^2\theta + \cos^2\theta = 1

and quadrant sign rules.

\sin\theta = -\sqrt{1 - \left(\frac{4}{5}\right)^2} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}

because sine is negative in Quadrant IV.

The Pythagorean identity relates sine and cosine, and the angle's quadrant determines the sign of the trigonometric ratio.

2

Calculate the values of

\tan\theta

,

\sec\theta

, and

\csc\theta

using reciprocal and quotient identities.

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-3/5}{4/5} = -\frac{3}{4}

,

\sec\theta = \frac{1}{\cos\theta} = \frac{5}{4}

, and

\csc\theta = \frac{1}{\sin\theta} = -\frac{5}{3}

.

Fundamental quotient and reciprocal identities define these functions in terms of sine and cosine.

3

Substitute the trigonometric values into the given expression

\frac{\tan\theta + \sec\theta}{\csc\theta}

and simplify.

\frac{-\frac{3}{4} + \frac{5}{4}}{-\frac{5}{3}} = \frac{\frac{2}{4}}{-\frac{5}{3}} = \frac{\frac{1}{2}}{-\frac{5}{3}} = \frac{1}{2} \times \left(-\frac{3}{5}\right) = -\frac{3}{10}

.

Combining terms in the numerator and dividing by the fraction in the denominator yields the simplified value.

Key Concept

Pythagorean, quotient, and reciprocal trigonometric identities with quadrant-dependent signs

Question 51Question →

Match each transformed trigonometric function listed on the left with the correct description of its key graphical features listed on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The trigonometric functions correctly match their graphical features as follows:

f(x)

matches the description with period

\pi

and midline

y = 1

;

g(x)

matches the description with period

4\pi

and phase shift

2\pi

left;

h(x)

matches the description with period

\pi/3

and y-intercept

(0, 3)

;

k(x)

matches the description with period

2

and midline

y = -3

.

Each trigonometric equation is mapped to its unique set of graphical properties by evaluating its period, midline, phase shift, and specific points like y-intercepts or extrema using standard trigonometric transformation formulas.

Step-by-Step Solution

1

Analyze

f(x) = -3\cos\left(2x - \frac{\pi}{2}\right) + 1

.

Factor out the coefficient of

x

:

f(x) = -3\cos\left(2\left(x - \frac{\pi}{4}\right)\right) + 1

. The period is

\frac{2\pi}{B} = \frac{2\pi}{2} = \pi

. The midline is

y = D = 1

. The maximum value occurs when the cosine term equals

-1

(due to the

-3

coefficient):

-3(-1) + 1 = 4

, which happens at

2x - \frac{\pi}{2} = \pi \implies x = \frac{3\pi}{4}

.

Identify period, midline, phase shift, and extrema from standard form

y = A\cos(B(x-C)) + D

.

2

Analyze

g(x) = 2\sin\left(\frac{1}{2}x + \pi\right) - 1

.

Rewrite as

g(x) = 2\sin\left(\frac{1}{2}(x - (-2\pi))\right) - 1

. The period is

\frac{2\pi}{1/2} = 4\pi

, and the phase shift is

2\pi

units to the left. The minimum value is

2(-1) - 1 = -3

, which occurs when

\frac{1}{2}x + \pi = \frac{3\pi}{2} \implies x = \pi

.

Determine horizontal shift, period, and minimum location.

3

Analyze

h(x) = -\tan\left(3x + \frac{3\pi}{4}\right) + 2

.

The period for tangent is

\frac{\pi}{B} = \frac{\pi}{3}

. Consecutive vertical asymptotes occur every period

\frac{\pi}{3}

. Evaluating at

x = 0

gives

h(0) = -\tan\left(\frac{3\pi}{4}\right) + 2 = -(-1) + 2 = 3

, giving a y-intercept of

(0, 3)

.

Apply tangent period formula

\frac{\pi}{|B|}

and evaluate y-intercept.

4

Analyze

k(x) = 4\cos\left(\pi x - \frac{\pi}{2}\right) - 3

.

Rewrite as

k(x) = 4\cos\left(\pi\left(x - \frac{1}{2}\right)\right) - 3

. The period is

\frac{2\pi}{\pi} = 2

. The phase shift is

\frac{1}{2}

unit to the right, and the midline is

y = -3

.

Extract parameters from cosine function with

\pi

in argument.

Key Concept

Graphical transformations of trigonometric functions (amplitude, period

T = \frac{2\pi}{|B|}

or

\frac{\pi}{|B|}

, phase shift

C

, and midline

D

).

Estimated Time:2m 0s

Question 52Question →

In right triangle $PQR$ , the right angle is at vertex $Q$ . Point $S$ lies on side $QR$ such that line segment $PS$ bisects $\angle QPR$ . If the length of $PQ$ is $14$ units and $\tan(\angle QPS) = \frac{3}{4}$ , what is the length, in units, of side $PR$ ?

17.5

28

35

50

56

Show answer & explanation

Answer: $50$

Answer

The length of side

PR

is

50

units.

In right triangle

PQS

, the tangent ratio gives

QS = 14 \cdot \frac{3}{4} = 10.5

, which yields

\sin(\angle QPS) = \frac{3}{5}

and

\cos(\angle QPS) = \frac{4}{5}

. Because

PS

bisects

\angle QPR

, the angle at

P

for triangle

PQR

is twice

\angle QPS

. Using the cosine double-angle relationship,

\cos(\angle QPR) = \cos^2(\angle QPS) - \sin^2(\angle QPS) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}

. Finally, applying SOHCAHTOA to right triangle

PQR

gives

\cos(\angle QPR) = \frac{PQ}{PR} = \frac{14}{PR} = \frac{7}{25}

, which solves to

PR = 50

.

Step-by-Step Solution

1

Use right triangle

PQS

to find

QS

and the trigonometric values of

\theta = \angle QPS

.

QS = 10.5

,

\sin(\theta) = \frac{3}{5}

, and

\cos(\theta) = \frac{4}{5}

.

Since

\triangle PQS

has a right angle at

Q

,

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{QS}{14} = \frac{3}{4}

, giving

QS = 10.5

. The hypotenuse

PS = \sqrt{14^2 + 10.5^2} = 17.5

, so

\sin(\theta) = \frac{10.5}{17.5} = \frac{3}{5}

and

\cos(\theta) = \frac{14}{17.5} = \frac{4}{5}

.

2

Determine

\cos(\angle QPR)

using the double-angle identity for cosine.

\cos(\angle QPR) = \frac{7}{25}.

Because

PS

bisects

\angle QPR

,

\angle QPR = 2\theta

. Using

\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)

, we get

\cos(2\theta) = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}

.

3

Apply the cosine definition SOHCAHTOA in right triangle

PQR

to solve for hypotenuse

PR

.

PR = 50.

In right triangle

PQR

,

\cos(\angle QPR) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{PQ}{PR} = \frac{14}{PR}

. Setting

\frac{14}{PR} = \frac{7}{25}

yields

7 \cdot PR = 350

, so

PR = 50

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Trigonometric Ratios of Composite Angles

Question 53Question →

The vertical displacement, $d(t)$ in centimeters, of a particle executing simple harmonic motion is modeled by the trigonometric function $d(t) = A \sin(B(t - C)) + D$ , where $A > 0$ , $B > 0$ , and $C$ represents the smallest non-negative phase shift in seconds. The graph of $d(t)$ completes one full cycle every $\frac{2\pi}{3}$ seconds, has a maximum value of $7\text{ cm}$ at $t = \frac{5\pi}{18}\text{ seconds}$ , and has a minimum value of $-3\text{ cm}$ . What is the value of $C$ ?

\frac{\pi}{18}

\frac{\pi}{9}

\frac{2\pi}{9}

\frac{\pi}{3}

\frac{5\pi}{18}

Show answer & explanation

Answer: $\frac{\pi}{9}$

Answer

\frac{\pi}{9}

To find the phase shift

C

, first determine

B

using the period formula

\text{Period} = \frac{2\pi}{B}

. Since the period is

\frac{2\pi}{3}

,

B = 3

. The function achieves a maximum when its sine argument equals

\frac{\pi}{2} + 2k\pi

. Setting

3\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2}

yields

\frac{5\pi}{6} - 3C = \frac{\pi}{2}

, which simplifies to

3C = \frac{\pi}{3}

, giving

C = \frac{\pi}{9}

.

Step-by-Step Solution

1

Determine the value of

B

from the period of the function.

B = 3

The standard period for a sine function is

2\pi

. Given that the period is

\frac{2\pi}{3}

, we set

\frac{2\pi}{B} = \frac{2\pi}{3}

, which yields

B = 3

.

2

Determine the maximum value equation for the parent sine function.

Argument equals

\frac{\pi}{2}

The standard sine function

\sin(\theta)

achieves its first positive maximum at

\theta = \frac{\pi}{2}

. Thus, for

d(t)

, the maximum occurs when

B(t - C) = \frac{\pi}{2}

.

3

Substitute

B = 3

and

t = \frac{5\pi}{18}

into the argument equation and solve for

C

.

C = \frac{\pi}{9}

Substitute the given values:

3\left(\frac{5\pi}{18} - C\right) = \frac{\pi}{2} \implies \frac{5\pi}{6} - 3C = \frac{\pi}{2}

. Subtract

\frac{5\pi}{6}

from both sides to get

-3C = \frac{3\pi}{6} - \frac{5\pi}{6} = -\frac{2\pi}{6} = -\frac{\pi}{3}

. Dividing by

-3

gives

C = \frac{\pi}{9}

.

Key Concept

Phase shift and parameter identification from graphs of transformed sine functions

Question 54Question →

If $\pi < \theta < \frac{3\pi}{2}$ and $\tan \theta = \frac{3}{4}$ , what is the value of the expression $\frac{\sin^2 \theta}{1 + \cos \theta}$ ?

\frac{9}{5}

\frac{1}{5}

\frac{8}{5}

-\frac{9}{5}

\frac{3}{5}

Show answer & explanation

Answer: $\frac{9}{5}$

Answer

\frac{9}{5}

The expression

\frac{\sin^2 \theta}{1 + \cos \theta}

can be simplified using the Pythagorean identity

\sin^2 \theta = 1 - \cos^2 \theta

. Factoring the numerator gives

(1 - \cos \theta)(1 + \cos \theta)

, which cancels with the denominator to leave

1 - \cos \theta

. Given

\tan \theta = \frac{3}{4}

in Quadrant III, the reference right triangle has sides 3 and 4 with hypotenuse 5. Since cosine is negative in the third quadrant,

\cos \theta = -\frac{4}{5}

. Evaluating

1 - \left(-\frac{4}{5}\right)

yields

\frac{9}{5}

.

Step-by-Step Solution

1

Simplify the algebraic trigonometric expression using standard fundamental identities.

\frac{\sin^2 \theta}{1 + \cos \theta} = \frac{1 - \cos^2 \theta}{1 + \cos \theta} = \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} = 1 - \cos \theta

Applying the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

allows factoring and canceling terms.

2

Determine the value and sign of

\cos \theta

based on the given tangent ratio and quadrant constraint.

\cos \theta = -\frac{4}{5}

Since

\tan \theta = \frac{3}{4} = \frac{\text{opposite}}{\text{adjacent}}

, the hypotenuse is

5

. In Quadrant III (

\pi < \theta < \frac{3\pi}{2}

), cosine is negative.

3

Substitute the value of

\cos \theta

into the simplified expression.

1 - \left(-\frac{4}{5}\right) = 1 + \frac{4}{5} = \frac{9}{5}

Subtracting a negative value results in addition.

Key Concept

Fundamental Pythagorean Identities and Quadrant Signs of Trigonometric Functions

Estimated Time:1m 15s

Question 55Question →

If $\sin \theta - \cos \theta = 0.6$ , what is the value of $\sin \theta \cos \theta$ ?

Show answer & explanation

Answer: 0.32

Answer

The value of

\sin \theta \cos \theta

is 0.32.

Squaring both sides of

\sin \theta - \cos \cos \theta = 0.6

yields

\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36

. Substituting the Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

gives

1 - 2\sin \theta \cos \theta = 0.36

, which rearranges to

2\sin \theta \cos \theta = 0.64

. Dividing by 2 yields

\sin \theta \cos \theta = 0.32

.

Step-by-Step Solution

1

Square both sides of the given equation

(\sin \theta - \cos \theta)^2 = 0.36

Squaring allows us to introduce the product

\sin \theta \cos \theta

alongside

\sin^2 \theta

and

\cos^2 \theta

.

2

Expand the binomial on the left side

\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36

Use the algebraic expansion identity

(a - b)^2 = a^2 - 2ab + b^2

.

3

Substitute the fundamental Pythagorean trigonometric identity

\sin^2 \theta + \cos^2 \theta = 1

1 - 2\sin \theta \cos \theta = 0.36

\sin^2 \theta + \cos^2 \theta

always equals 1 for any angle

\theta

.

4

Isolate the term containing

\sin \theta \cos \theta

2\sin \theta \cos \theta = 0.64

Subtract 0.36 from 1 to find the value of

2\sin \theta \cos \theta

.

5

Divide by 2 to solve for

\sin \theta \cos \theta

\sin \theta \cos \theta = 0.32

Simplifies

0.64 / 2

to obtain the final required numerical value.

Key Concept

Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

Question 56Question →

A triangular region $PQR$ has side lengths $PQ = 8$ meters and $QR = 10$ meters, and the measure of angle $\angle PQR$ is $120^\circ$ . A straight walkway $QM$ is constructed from vertex $Q$ to a point $M$ on side $PR$ such that $QM$ bisects $\angle PQR$ . What is the length, in meters, of the walkway $QM$ ?

\frac{40}{9}

\frac{80}{9}

\frac{20}{9}

\frac{40}{3}

4\sqrt{3}

Show answer & explanation

Answer: $\frac{40}{9}$

Answer

\frac{40}{9}

meters

The total area of triangle

PQR

is equal to the sum of the areas of triangles

PQM

and

MQR

. Using the formula

\text{Area} = \frac{1}{2}ab\sin(\theta)

, the equation

\frac{1}{2}(8)(10)\sin(120^\circ) = \frac{1}{2}(8)(x)\sin(60^\circ) + \frac{1}{2}(10)(x)\sin(60^\circ)

simplifies to

20\sqrt{3} = \frac{9x\sqrt{3}}{2}

, which gives

x = \frac{40}{9}

.

Step-by-Step Solution

1

Express the area of the entire triangle

PQR

using the sine area formula.

\text{Area}(\triangle PQR) = \frac{1}{2} \cdot PQ \cdot QR \cdot \sin(120^\circ) = \frac{1}{2} \cdot 8 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}

square meters.

The area of a triangle given two sides and the included angle is

\frac{1}{2}ab\sin(C)

.

2

Express the sum of the areas of the two smaller triangles

\triangle PQM

and

\triangle MQR

created by the angle bisector

QM = x

.

Since

QM

bisects

\angle PQR = 120^\circ

,

\angle PQM = 60^\circ

and

\angle MQR = 60^\circ

.

\text{Area}(\triangle PQM) = \frac{1}{2} \cdot 8 \cdot x \cdot \sin(60^\circ) = 2x\sqrt{3}

.

\text{Area}(\triangle MQR) = \frac{1}{2} \cdot 10 \cdot x \cdot \sin(60^\circ) = \frac{5x\sqrt{3}}{2}

.

An angle bisector divides the total angle into two equal

60^\circ

angles.

3

Equate the total area to the sum of the partial areas and solve for

x

.

20\sqrt{3} = 2x\sqrt{3} + \frac{5x\sqrt{3}}{2} \implies 20 = 2x + \frac{5x}{2} \implies 20 = \frac{9x}{2} \implies x = \frac{40}{9}

.

The total area of the figure is equal to the sum of its non-overlapping component areas.

Key Concept

Area of Triangles using Sine and Angle Bisector Properties

Question 57Question →

In right triangle $JKL$ , the right angle is at vertex $K$ . Line segment $KM$ is perpendicular to hypotenuse $JL$ , with point $M$ lying on $JL$ . If $\sin(\angle KJL) = \frac{4}{5}$ and the length of segment $JM$ is $9$ units, what is the length, in units, of side $KL$ ?

9

12

15

20

25

Show answer & explanation

Answer: $20$

Answer

20

By using SOHCAHTOA,

\sin(\angle KJL) = \frac{4}{5}

implies that

\cos(\angle KJL) = \frac{3}{5}

and

\tan(\angle KJL) = \frac{4}{3}

. In the smaller right triangle

\triangle JMK

,

\cos(\angle KJL) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{JM}{JK}

, so

\frac{3}{5} = \frac{9}{JK}

, which yields

JK = 15

. Next, in the large right triangle

\triangle JKL

,

\tan(\angle KJL) = \frac{\text{opposite}}{\text{adjacent}} = \frac{KL}{JK}

, so

\frac{4}{3} = \frac{KL}{15}

, giving

KL = 20

.

Step-by-Step Solution

1

Determine

\cos(\angle KJL)

and

\tan(\angle KJL)

using SOHCAHTOA.

\cos(\angle KJL) = \frac{3}{5}

and

\tan(\angle KJL) = \frac{4}{3}

Since

\sin(\angle KJL) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}

in a right triangle, the adjacent side ratio is

\sqrt{5^2 - 4^2} = 3

, giving

\cos(\angle KJL) = \frac{3}{5}

and

\tan(\angle KJL) = \frac{4}{3}

.

2

Apply the cosine ratio in right triangle

\triangle JMK

(where

\angle JMK = 90^\circ

).

JK = 15

In

\triangle JMK

,

\cos(\angle KJL) = \frac{JM}{JK}

. Substituting the given values gives

\frac{3}{5} = \frac{9}{JK} \implies 3 \cdot JK = 45 \implies JK = 15

.

3

Apply the tangent ratio in main right triangle

\triangle JKL

to calculate

KL

.

KL = 20

In

\triangle JKL

,

\tan(\angle KJL) = \frac{KL}{JK}

. Substituting

JK = 15

gives

\frac{4}{3} = \frac{KL}{15} \implies 3 \cdot KL = 60 \implies KL = 20

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) across Similar Right Triangles

Estimated Time:2m 0s

Question 58Question →

In triangular plot $ABC$ , the boundary lengths are $AB = 13$ meters, $BC = 8$ meters, and $AC = 15$ meters. A straight drainage pipe is laid from vertex $B$ perpendicular to side $AC$ , meeting side $AC$ at point $D$ . What is the distance, in meters, from point $A$ to point $D$ ?

Show answer & explanation

Answer: 11

Answer

The distance from point A to point D is 11 meters.

Applying the Law of Cosines

a^2 = b^2 + c^2 - 2bc \cos A

with

a=8

,

b=15

, and

c=13

yields

64 = 225 + 169 - 390 \cos A

, which simplifies to

390 \cos A = 330

, giving

\cos A = \frac{11}{13}

. In right triangle

ABD

,

\cos A = \frac{AD}{AB}

, so

AD = 13 \cdot \frac{11}{13} = 11

meters.

Step-by-Step Solution

1

Apply the Law of Cosines to triangle ABC to solve for the cosine of angle A.

cos A = 11/13

The Law of Cosines relates all three side lengths of a triangle to the cosine of one of its interior angles.

2

Use right triangle trigonometry in right triangle ABD to calculate the length of AD.

AD = 11 meters

Since BD is perpendicular to AC, triangle ABD is a right triangle with hypotenuse AB and adjacent side AD relative to angle A.

Key Concept

Law of Cosines and Right Triangle Trigonometry

Estimated Time:2m 0s

Question 59Question →

A surveyor stands at point $A$ on horizontal ground and measures the angle of elevation to the top of a vertical cliff, point $C$ , such that $\tan(\angle CAD) = \frac{1}{2}$ , where $D$ is the base of the cliff directly below $C$ . The surveyor then walks $50$ feet closer to the cliff along a straight horizontal path to point $B$ , where the angle of elevation to point $C$ satisfies $\tan(\angle CBD) = \frac{4}{3}$ . Points $A$ , $B$ , and $D$ are collinear. What is the height, in feet, of the cliff?

25

30

40

60

75

Show answer & explanation

Answer: 40

Answer

40 feet

By applying SOHCAHTOA to both right triangles, we set up

\tan(\angle CBD) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{BD} = \frac{4}{3}

, giving

BD = \frac{3}{4}h

. For the larger triangle,

\tan(\angle CAD) = \frac{h}{50 + BD} = \frac{1}{2}

. Substituting

BD = \frac{3}{4}h

into the equation gives

50 + \frac{3}{4}h = 2h

, which solves directly to

h = 40

feet.

Step-by-Step Solution

1

Define the unknown quantities using the right triangles formed by the cliff and the observation points.

Let

h = CD

be the height of the cliff, and let

d = BD

be the horizontal distance from point

B

to the cliff base

D

. The total distance from

A

to

D

is

AD = AB + BD = 50 + d

.

Establishing explicit variables allows us to translate the geometric relationships into algebraic equations.

2

Apply the tangent ratio (SOHCAHTOA:

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}

) to right triangle

BCD

.

\tan(\angle CBD) = \frac{CD}{BD} \implies \frac{4}{3} = \frac{h}{d} \implies d = \frac{3}{4}h

Expressing the distance

d

in terms of height

h

enables substitution into the second right triangle equation.

3

Apply the tangent ratio to right triangle

ACD

and substitute

d = \frac{3}{4}h

.

\tan(\angle CAD) = \frac{CD}{AD} \implies \frac{1}{2} = \frac{h}{50 + d} \implies \frac{1}{2} = \frac{h}{50 + \frac{3}{4}h}

This creates a single linear equation in terms of the cliff height

h

.

4

Solve the equation for

h

.

50 + \frac{3}{4}h = 2h \implies 50 = 2h - \frac{3}{4}h \implies 50 = \frac{5}{4}h \implies h = 40

Cross-multiplying and isolating

h

yields the correct height of the cliff in feet.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:1m 30s

Question 60Question →

For an angle $\theta$ in the third quadrant satisfying $\pi < \theta < \frac{3\pi}{2}$ , the tangent value is $\tan \theta = \frac{4}{3}$ . What is the exact value of the expression $\sin^4 \theta - \cos^4 \theta$ ?

Show answer & explanation

Answer: 0.28

Answer

The exact numerical value of the expression is 0.28 (or 7/25).

By factoring

\sin^4 \theta - \cos^4 \theta

as

(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)

, we can apply the fundamental Pythagorean identity

\sin^2 \theta + \cos^2 \theta = 1

. The expression simplifies cleanly to

\sin^2 \theta - \cos^2 \theta

. Given

\tan \theta = \frac{4}{3}

in Quadrant III, the reference triangle has opposite side 4, adjacent side 3, and hypotenuse 5. Thus,

\sin \theta = -\frac{4}{5}

and

\cos \theta = -\frac{3}{5}

. Substituting these values yields

\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28

.

Step-by-Step Solution

1

Factor the fourth-degree trigonometric expression using difference of squares.

\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)

The difference of two squares

a^2 - b^2 = (a-b)(a+b)

applies directly to

a = \sin^2 \theta

and

b = \cos^2 \theta

.

2

Simplify using the fundamental Pythagorean trigonometric identity.

\sin^4 \theta - \cos^4 \theta = \sin^2 \theta - \cos^2 \theta

By the Pythagorean identity,

\sin^2 \theta + \cos^2 \theta = 1

for any angle

\theta

.

3

Determine

\sin \theta

and

\cos \theta

from the given quadrant and tangent value.

\sin \theta = -\frac{4}{5}

and

\cos \theta = -\frac{3}{5}

In Quadrant III (

\pi < \theta < \frac{3\pi}{2}

), both sine and cosine are negative. A standard 3-4-5 right triangle yields

\sin \theta = -\frac{4}{5}

and

\cos \theta = -\frac{3}{5}

.

4

Substitute the values into the simplified expression and compute the result.

\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} = 0.28

Squaring each trigonometric ratio yields positive values, giving a final simplified decimal value of 0.28.

Key Concept

Pythagorean Identity and Difference of Squares

Estimated Time:1m 30s

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