Trigonometry Practice Questions — ACT — Page 5

Question 81Question →

A park planner is designing a triangular walking trail connecting three landmarks: a fountain at point $F$ , a gazebo at point $G$ , and a bridge at point $B$ . The distance from the fountain to the gazebo is $800$ meters, and the distance from the gazebo to the bridge is $1{}500$ meters. If the angle formed at the gazebo ( $\angle FGB$ ) measures $60^\circ$ , what is the direct distance, in meters, from the fountain at point $F$ to the bridge at point $B$ ?

Show answer & explanation

Answer: 1300

Answer

1300 meters

Applying the Law of Cosines

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

with

a = 800

,

b = 1500

, and

C = 60^\circ

gives

c^2 = 800^2 + 1500^2 - 2(800)(1500)(0.5) = 1{}690{}000

. Taking the square root yields

c = 1300

meters.

Step-by-Step Solution

1

Identify given measurements and formula

FG = 800

m,

GB = 1500

m,

\angle FGB = 60^\circ

, using Law of Cosines

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

Two side lengths and the included angle (SAS) are known, requiring the Law of Cosines to solve for the opposite side.

2

Substitute values into the Law of Cosines equation

FB^2 = 800^2 + 1500^2 - 2(800)(1500)\cos(60^\circ)

Direct replacement of known side lengths and angle measure into the formula.

3

Evaluate the arithmetic terms

FB^2 = 640{}000 + 2{}250{}000 - 1{}200{}000 = 1{}690{}000

Squaring the side lengths and calculating the product

2(800)(1500)(0.5)

.

4

Take the square root to solve for

FB

FB = \sqrt{1{}690{}000} = 1300

Taking the principal square root yields the distance in meters.

Key Concept

Law of Cosines (SAS Triangle Solving)

Question 82Question →

A drone is positioned in the sky above a flat field between two ground observation stations, Station $A$ and Station $B$ , which are $500$ meters apart. From Station $A$ , the angle of elevation to the drone is $40^\circ$ . From Station $B$ , looking back toward the drone, the angle of elevation is $65^\circ$ . Assuming the drone and both observation stations lie in the same vertical plane, which of the following expressions represents the direct distance, in meters, from Station $A$ to the drone?

\frac{500 \sin(40^\circ)}{\sin(75^\circ)}

\frac{500 \sin(65^\circ)}{\sin(75^\circ)}

\frac{500 \sin(75^\circ)}{\sin(65^\circ)}

\frac{500 \sin(65^\circ)}{\sin(40^\circ)}

\frac{500 \sin(40^\circ)}{\sin(65^\circ)}

Show answer & explanation

Answer: $\frac{500 \sin(65^\circ)}{\sin(75^\circ)}$

Answer

The correct expression is

\frac{500 \sin(65^\circ)}{\sin(75^\circ)}

.

To find the distance from Station

A

to the drone (

AD

), model the scenario as triangle

ABD

. The ground side

AB = 500

meters. The interior angle at

A

is

40^\circ

and at

B

is

65^\circ

. The third angle at the drone

D

is

180^\circ - 40^\circ - 65^\circ = 75^\circ

. By the Law of Sines,

\frac{AD}{\sin(65^\circ)} = \frac{500}{\sin(75^\circ)}

, which simplifies to

AD = \frac{500 \sin(65^\circ)}{\sin(75^\circ)}

.

Step-by-Step Solution

1

Determine the interior angle of the triangle at the drone's location (

D

).

\angle D = 180^\circ - (40^\circ + 65^\circ) = 75^\circ

The interior angles of any triangle must sum to

180^\circ

.

2

Apply the Law of Sines to relate the known side

AB = 500

meters and its opposite angle

\angle D = 75^\circ

to the unknown side

AD

and its opposite angle

\angle B = 65^\circ

.

\frac{AD}{\sin(65^\circ)} = \frac{500}{\sin(75^\circ)}

The Law of Sines states that in any triangle,

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

.

3

Solve for the side length

AD

.

AD = \frac{500 \sin(65^\circ)}{\sin(75^\circ)}

Multiplying both sides of the equation by

\sin(65^\circ)

isolates

AD

.

Key Concept

Applying the Law of Sines to find an unknown side length given two angles and one side of a triangle.

Estimated Time:1m 15s

Question 83Question →

In right triangle $JKM$ , the right angle is at vertex $K$ . Leg $JK$ has a length of $16$ centimeters. If $\sin(\angle M) = \frac{8}{17}$ , what is the length, in centimeters, of leg $KM$ ?

15

30

34

2\sqrt{353}

\frac{136}{15}

Show answer & explanation

Answer: $30$

Answer

The length of leg

KM

is

30

centimeters.

By definition of sine in a right triangle,

\sin(\angle M) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{JK}{JM}

. Given

\sin(\angle M) = \frac{8}{17}

and

JK = 16

, setting

\frac{8}{17} = \frac{16}{JM}

yields hypotenuse

JM = 34\text{ cm}

. Using the Pythagorean theorem

KM = \sqrt{JM^2 - JK^2} = \sqrt{34^2 - 16^2} = \sqrt{900} = 30\text{ cm}

. Therefore, the option with value

30

is correct.

Step-by-Step Solution

1

Use the definition of sine to find the hypotenuse

JM

.

\sin(\angle M) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{JK}{JM} \implies \frac{8}{17} = \frac{16}{JM} \implies JM = 34\text{ cm}

.

Sine is defined as the ratio of the side opposite the angle to the hypotenuse in a right triangle.

2

Apply the Pythagorean theorem to solve for the missing leg

KM

.

JK^2 + KM^2 = JM^2 \implies 16^2 + KM^2 = 34^2 \implies 256 + KM^2 = 1156 \implies KM^2 = 900 \implies KM = 30\text{ cm}

.

In any right triangle, the sum of the squares of the legs equals the square of the hypotenuse.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Question 84Question →

A 25-foot ladder leans against a vertical wall on flat, horizontal ground. The base of the ladder is positioned 7 feet away from the wall. If $\theta$ represents the measure of the angle formed between the ladder and the ground, what is the value of $\sin(\theta) + \cos(\theta)$ ?

\frac{31}{25}

\frac{17}{25}

\frac{31}{24}

\frac{31}{7}

\frac{31}{\sqrt{674}}

Show answer & explanation

Answer: $\frac{31}{25}$

Answer

\frac{31}{25}

The ladder forms a 7-24-25 right triangle with the ground and the wall. Relative to angle

\theta

between the ladder and the ground, the opposite side is 24 feet, the adjacent side is 7 feet, and the hypotenuse is 25 feet. Evaluating

\sin(\theta) = \frac{24}{25}

and

\cos(\theta) = \frac{7}{25}

and adding them gives

\frac{31}{25}

.

Step-by-Step Solution

1

Identify known components of the right triangle

The ladder acts as the hypotenuse (

c = 25

), and the distance along the ground is the adjacent side to angle

\theta

(

a = 7

).

The ladder, wall, and ground form a right triangle with the right angle at the base of the wall.

2

Calculate the height of the wall (opposite side)

Opposite side

b = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24

feet.

By the Pythagorean theorem (

a^2 + b^2 = c^2

), solving for

b

gives

b = \sqrt{c^2 - a^2}

.

3

Determine

\sin(\theta)

and

\cos(\theta)

using SOHCAHTOA ratios

\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{24}{25}

and

\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{7}{25}

.

Sine is defined as opposite over hypotenuse, and cosine is defined as adjacent over hypotenuse.

4

Sum the trigonometric ratios

\sin(\theta) + \cos(\theta) = \frac{24}{25} + \frac{7}{25} = \frac{31}{25}

.

Add the two fractions with the common denominator 25.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Theorem

Estimated Time:1m 15s

Question 85Question →

An architect is designing a triangular solar panel frame with side lengths measuring $15$ feet, $24$ feet, and $21$ feet. What is the measure, in degrees, of the interior angle opposite the side measuring $21$ feet?

Show answer & explanation

Answer: 60

Answer

The measure of the interior angle opposite the side measuring 21 feet is 60 degrees.

Using the Law of Cosines

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

with side lengths

a = 15

,

b = 24

, and

c = 21

gives

21^2 = 15^2 + 24^2 - 2(15)(24)\cos(C)

. Simplifying the equation leads to

441 = 801 - 720\cos(C)

, which rearranges to

-360 = -720\cos(C)

or

\cos(C) = 0.5

. Evaluating

\arccos(0.5)

yields an angle measure of

60^\circ

.

Step-by-Step Solution

1

Set up the Law of Cosines with the side lengths

a = 15

,

b = 24

, and target opposite side

c = 21

.

21^2 = 15^2 + 24^2 - 2(15)(24)\cos(C)

The Law of Cosines connects three side lengths of any triangle to the cosine of the angle opposite one of those sides.

2

Simplify the numerical values in the equation.

441 = 801 - 720\cos(C)

Evaluate

21^2 = 441

,

15^2 + 24^2 = 225 + 576 = 801

, and

2(15)(24) = 720

.

3

Isolate the cosine expression.

\cos(C) = 0.5

Subtracting

801

from both sides gives

-360 = -720\cos(C)

, then dividing by

-720

yields

0.5

.

4

Find the inverse cosine of

0.5

.

C = 60^\circ

In any triangle, the angle whose cosine is

0.5

is

60^\circ

.

Key Concept

Applying the Law of Cosines to solve for an unknown angle given all three side lengths of a non-right triangle.

Question 86Question →

In rhombus $ABCD$ , the diagonals $AC$ and $BD$ intersect at point $E$ . If the length of diagonal $AC$ is $16$ centimeters and the length of side $AB$ is $10$ centimeters, what is the value of $\sin(\angle ABE)$ ?

\frac{3}{5}

\frac{4}{5}

\frac{3}{4}

\frac{4}{3}

\frac{5}{4}

Show answer & explanation

Answer: $\frac{4}{5}$

Answer

The value of

\sin(\angle ABE)

is

\frac{4}{5}

.

The correct answer is

\frac{4}{5}

. Because the diagonals of a rhombus are perpendicular bisectors,

\triangle ABE

is a right triangle with the right angle at vertex

E

. Diagonal

AC

is bisected at

E

, making

AE = 8

cm. For

\angle ABE

, the opposite side is

AE = 8

cm and the hypotenuse is

AB = 10

cm. By SOHCAHTOA,

\sin(\angle ABE) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{8}{10} = \frac{4}{5}

.

Step-by-Step Solution

1

Use the properties of a rhombus to identify right triangles and segment lengths.

The diagonals of a rhombus intersect at right angles (

90^\circ

) and bisect each other. Therefore,

\triangle ABE

is a right triangle with right angle at vertex

E

, and leg

AE = \frac{1}{2} AC = \frac{1}{2}(16) = 8

cm.

Rhombus diagonals are perpendicular bisectors.

2

Identify the sides of right triangle

ABE

relative to angle

\angle ABE

.

The hypotenuse is side

AB = 10

cm, and the side opposite to

\angle ABE

is leg

AE = 8

cm.

Opposite side is directly across from the angle of interest.

3

Apply the sine ratio formula (SOHCAHTOA).

\sin(\angle ABE) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AE}{AB} = \frac{8}{10} = \frac{4}{5}

Sine is defined as the ratio of the opposite side length to the hypotenuse length.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) applied to Rhombus Geometry

Estimated Time:1m 15s

Question 87Question →

A sailboat is traveling along a straight path. From point $A$ , a lighthouse $L$ is observed at an angle of $35^\circ$ relative to the line of travel. After the boat travels $400$ meters directly along the path to reach point $C$ , the angle to the lighthouse relative to the continuing line of travel increases to $65^\circ$ . Which of the following expressions represents the distance, in meters, from point $C$ to the lighthouse $L$ ?

800 \sin(35^\circ)

400 \sin(35^\circ)

\frac{400 \sin(35^\circ)}{\sin(65^\circ)}

\frac{400 \sin(65^\circ)}{\sin(35^\circ)}

800 \sin(65^\circ)

Show answer & explanation

Answer: $800 \sin(35^\circ)$

Answer

800 \sin(35^\circ)

meters

To find the distance from

C

to the lighthouse

L

, analyze

\triangle ALC

. The given angle at

A

is

35^\circ

. Point

C

is along the straight line of travel, so the interior angle

\angle LCA = 180^\circ - 65^\circ = 115^\circ

. The top angle

\angle ALC = 180^\circ - (35^\circ + 115^\circ) = 30^\circ

. By the Law of Sines,

\frac{CL}{\sin(35^\circ)} = \frac{400}{\sin(30^\circ)}

. Since

\sin(30^\circ) = 0.5

, solving for

CL

gives

CL = \frac{400 \sin(35^\circ)}{0.5} = 800 \sin(35^\circ)

.

Step-by-Step Solution

1

Determine the interior angles of

\triangle ALC

Angle

\angle LAC = 35^\circ

. The supplementary interior angle at

C

is

\angle LCA = 180^\circ - 65^\circ = 115^\circ

. The third interior angle

\angle ALC = 180^\circ - (35^\circ + 115^\circ) = 30^\circ

.

The interior angle and exterior angle along a straight path sum to

180^\circ

, and the interior angles of a triangle sum to

180^\circ

.

2

Apply the Law of Sines to find distance

CL

\frac{CL}{\sin(35^\circ)} = \frac{AC}{\sin(30^\circ)} \implies \frac{CL}{\sin(35^\circ)} = \frac{400}{\sin(30^\circ)}

The Law of Sines states that the ratio of a side length to the sine of its opposite angle is constant in any triangle.

3

Simplify the expression using

\sin(30^\circ) = 0.5

CL = \frac{400 \sin(35^\circ)}{0.5} = 800 \sin(35^\circ)

Dividing

400

by

0.5

yields

800

.

Key Concept

Law of Sines

Question 88Question →

Two radar stations, $A$ and $B$ , are located $12$ miles apart along a straight coastline. Both stations track a ship offshore at point $C$ . The angle formed by the coastline $AB$ and the line of sight from station $A$ to the ship ( $\angle BAC$ ) measures $42^\circ$ , and the angle formed by the coastline $AB$ and the line of sight from station $B$ to the ship ( $\angle ABC$ ) measures $78^\circ$ . Which of the following expressions represents the distance, in miles, from station $A$ to the ship?

\frac{12 \sin(42^\circ)}{\sin(60^\circ)}

\frac{12 \sin(78^\circ)}{\sin(60^\circ)}

\frac{12 \sin(60^\circ)}{\sin(78^\circ)}

\frac{12 \sin(78^\circ)}{\sin(42^\circ)}

\frac{12 \sin(42^\circ)}{\sin(78^\circ)}

Show answer & explanation

Answer: $\frac{12 \sin(78^\circ)}{\sin(60^\circ)}$

Answer

The distance, in miles, from station A to the ship is given by

\frac{12 \sin(78^\circ)}{\sin(60^\circ)}

.

The sum of angles in

\triangle ABC

is

180^\circ

, so

\angle C = 180^\circ - 42^\circ - 78^\circ = 60^\circ

. The distance from station A to the ship corresponds to side length

AC

, which lies opposite

\angle B = 78^\circ

. Applying the Law of Sines yields

\frac{AC}{\sin(78^\circ)} = \frac{12}{\sin(60^\circ)}

, which simplifies to

AC = \frac{12 \sin(78^\circ)}{\sin(60^\circ)}

.

Step-by-Step Solution

1

Calculate the measure of the third angle

\angle ACB

in

\triangle ABC

.

\angle ACB = 180^\circ - (42^\circ + 78^\circ) = 60^\circ

The interior angles of any triangle must sum to

180^\circ

.

2

Set up the Law of Sines relating the known side

AB = 12

and its opposite angle

\angle ACB = 60^\circ

to the unknown side

AC = b

and its opposite angle

\angle ABC = 78^\circ

.

\frac{AC}{\sin(78^\circ)} = \frac{12}{\sin(60^\circ)}

The Law of Sines states that

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

.

3

Solve the equation for

AC

.

AC = \frac{12 \sin(78^\circ)}{\sin(60^\circ)}

Multiply both sides of the equation by

\sin(78^\circ)

.

Key Concept

Law of Sines

Estimated Time:1m 15s

Question 89Question →

A sailboat travels due east from Port $P$ for $15$ nautical miles to Point $Q$ , then turns due north and travels $36$ nautical miles to Point $R$ . The straight line of sight from Port $P$ to Point $R$ forms an angle $\theta$ with segment $PQ$ . What is the value of $\cos \theta - \sin \theta$ ?

-\frac{7}{13}

\frac{7}{13}

\frac{17}{13}

-\frac{7}{17}

-\frac{12}{5}

Show answer & explanation

Answer: $-\frac{7}{13}$

Answer

The correct value of

\cos \theta - \sin \theta

is

-\frac{7}{13}

.

The direct distance between Port

P

and Point

R

forms the hypotenuse of right triangle

PQR

. By applying the Pythagorean theorem, the hypotenuse length is

\sqrt{15^2 + 36^2} = 39

. Relative to angle

\theta

at vertex

P

, the adjacent leg is

15

and the opposite leg is

36

. Therefore,

\cos \theta = \frac{15}{39} = \frac{5}{13}

and

\sin \theta = \frac{36}{39} = \frac{12}{13}

. Subtracting these yields

\frac{5}{13} - \frac{12}{13} = -\frac{7}{13}

.

Step-by-Step Solution

1

Determine the length of the hypotenuse

PR

using the Pythagorean theorem.

PR = \sqrt{15^2 + 36^2} = \sqrt{225 + 1296} = \sqrt{1521} = 39

nautical miles.

Triangle

PQR

is a right triangle with right angle at vertex

Q

because the path turns from due east to due north.

2

Identify the opposite side, adjacent side, and hypotenuse relative to angle

\theta = \angle QPR

.

Adjacent side

PQ = 15

, Opposite side

QR = 36

, Hypotenuse

PR = 39

.

Angle

\theta

is formed at vertex

P

between the base path

PQ

and the direct distance

PR

.

3

Calculate

\cos \theta

and

\sin \theta

using SOHCAHTOA ratios.

\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{15}{39} = \frac{5}{13}

and

\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{36}{39} = \frac{12}{13}

.

Cosine is adjacent over hypotenuse and sine is opposite over hypotenuse.

4

Evaluate the expression

\cos \theta - \sin \theta

.

\cos \theta - \sin \theta = \frac{5}{13} - \frac{12}{13} = -\frac{7}{13}

.

Subtracting the sine value from the cosine value gives a negative result since the opposite leg is longer than the adjacent leg.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:1m 30s

Question 90Question →

A landscape architect is designing a triangular courtyard garden. Two adjacent edges of the garden measure $8$ meters and $15$ meters, and the angle between these two edges is $60^\circ$ . What is the length, in meters, of the third edge of the garden?

Show answer & explanation

Answer: 13

Answer

The length of the third edge of the garden is 13 meters.

Applying the Law of Cosines

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

with

a = 8

,

b = 15

, and

C = 60^\circ

yields

c^2 = 8^2 + 15^2 - 2(8)(15)\cos(60^\circ) = 64 + 225 - 240(0.5) = 169

. Taking the square root gives

c = 13

meters.

Step-by-Step Solution

1

Identify the given side lengths and included angle.

Two sides are

a = 8\text{ m}

and

b = 15\text{ m}

, and their included angle is

C = 60^\circ

.

The Law of Cosines is used when two sides and the included angle (SAS) are known.

2

Substitute the values into the Law of Cosines formula

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

.

c^2 = 8^2 + 15^2 - 2(8)(15)\cos(60^\circ) = 64 + 225 - 240(0.5) = 169

.

Evaluating the squared terms and trigonometric value

\cos(60^\circ) = 0.5

simplifies the equation to

c^2 = 169

.

3

Solve for the side length

c

by taking the square root.

c = \sqrt{169} = 13\text{ meters}

.

The physical side length of a geometric figure must be positive.

Key Concept

Law of Cosines (

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

)

Estimated Time:1m 30s

Question 91Question →

A civil engineer is designing a triangular bridge support structure with vertices $P$ , $Q$ , and $R$ . The support beam $PQ$ is $80$ feet long, the beam $QR$ is $50$ feet long, and the interior angle $\angle PQR$ measures $60^\circ$ . What is the length, in feet, of the support beam $PR$ ?

Show answer & explanation

Answer: 70

Answer

The length of the support beam

PR

is

70

feet.

Using the Law of Cosines formula

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

with side lengths

80

and

50

and included angle

60^\circ

, we calculate

PR^2 = 80^2 + 50^2 - 2(80)(50)\cos(60^\circ) = 6400 + 2500 - 4000 = 4900

. Taking the square root gives

PR = 70

feet.

Step-by-Step Solution

1

Identify the given dimensions and included angle

Side

PQ = 80

ft, side

QR = 50

ft, and included angle

\angle PQR = 60^\circ

.

The Law of Cosines applies directly when two side lengths and the included angle (SAS) are known.

2

Set up the Law of Cosines equation for the unknown side

PR

PR^2 = PQ^2 + QR^2 - 2(PQ)(QR)\cos(\angle PQR)

This formula generalizes the Pythagorean theorem to non-right triangles.

3

Substitute the known values into the equation and evaluate

PR^2 = 80^2 + 50^2 - 2(80)(50)\cos(60^\circ) = 6400 + 2500 - 4000 = 4900

Evaluating squares and using

\cos(60^\circ) = 0.5

simplifies the calculation.

4

Take the positive square root to find

PR

PR = \sqrt{4900} = 70

Side length must be a positive real number.

Key Concept

Applying the Law of Cosines to solve for an unknown side in a Side-Angle-Side (SAS) triangle.

Question 92Question →

A surveyor standing at point $A$ on level ground measures the angle of elevation to the top, point $C$ , of a vertical observation tower. The base of the tower is at point $B$ , forming right triangle $ABC$ with the right angle at vertex $B$ . The horizontal distance along the ground from point $A$ to base $B$ is $40$ meters. If $\sin(\angle A) = \frac{9}{41}$ , what is the height, in meters, of the tower?

9

\frac{360}{41}

40

41

\frac{160}{9}

Show answer & explanation

Answer: $9$

Answer

The height of the tower is

9

meters.

By definition of SOHCAHTOA,

\sin(\angle A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{9}{41}

. Since

\cos(\angle A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC}

, and using the Pythagorean triple

9

-

40

-

41

, we find

\cos(\angle A) = \frac{40}{41}

. Given that the adjacent side

AB = 40

meters, the hypotenuse

AC

must be

41

meters, which means the opposite side (tower height

BC

) is

9

meters.

Step-by-Step Solution

1

Identify the given trigonometric ratio and sides of the right triangle.

In right triangle

ABC

with right angle at

B

, the adjacent side to

\angle A

is

AB = 40

, the opposite side is

BC

(height of the tower), and the hypotenuse is

AC

. We are given

\sin(\angle A) = \frac{BC}{AC} = \frac{9}{41}

.

SOHCAHTOA defines sine as the ratio of the opposite side to the hypotenuse.

2

Determine the relationship between the sides using the Pythagorean theorem or cosine ratio.

Since

\sin(\angle A) = \frac{9}{41}

, we know

\cos(\angle A) = \frac{AB}{AC} = \sqrt{1 - \left(\frac{9}{41}\right)^2} = \sqrt{\frac{1681 - 81}{1681}} = \sqrt{\frac{1600}{1681}} = \frac{40}{41}

.

The Pythagorean identity

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

relates sine and cosine for any acute angle in a right triangle.

3

Calculate the length of the hypotenuse and the height of the tower.

Setting

\cos(\angle A) = \frac{40}{AC} = \frac{40}{41}

gives

AC = 41

meters. Then,

BC = 41 \times \sin(\angle A) = 41 \times \frac{9}{41} = 9

meters.

Substituting the known adjacent length of

40

meters into the ratio yields the exact vertical height.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Triples

Estimated Time:1m 0s

Question 93Question →

A straight ramp is constructed for a skateboard park. The ramp rises to a vertical height of $5\sqrt{3}$ feet above flat horizontal ground, and the horizontal distance from the base of the ramp to the point directly beneath its highest point is $15$ feet. If $\theta$ represents the angle of inclination of the ramp with respect to the ground, what is the value of $\cos(\theta)$ ?

\frac{1}{2}

\frac{\sqrt{3}}{3}

\frac{\sqrt{3}}{2}

\frac{2\sqrt{3}}{3}

\sqrt{3}

Show answer & explanation

Answer: $\frac{\sqrt{3}}{2}$

Answer

The value of

\cos(\theta)

is

\frac{\sqrt{3}}{2}

.

The horizontal distance of

15

feet is adjacent to angle

\theta

, and the vertical height of

5\sqrt{3}

feet is opposite to

\theta

. By the Pythagorean theorem, the hypotenuse is

\sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{225 + 75} = 10\sqrt{3}

feet. Using SOHCAHTOA,

\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{15}{10\sqrt{3}} = \frac{\sqrt{3}}{2}

.

Step-by-Step Solution

1

Identify the given side lengths relative to the angle

\theta

The side opposite to angle

\theta

is

5\sqrt{3}

feet, and the side adjacent to angle

\theta

is

15

feet.

The height of the ramp is opposite to the angle of inclination, and the horizontal ground distance is adjacent.

2

Calculate the length of the hypotenuse using the Pythagorean theorem

Hypotenuse

= \sqrt{15^2 + (5\sqrt{3})^2} = \sqrt{225 + 75} = \sqrt{300} = 10\sqrt{3}

feet.

Cosine is defined as adjacent divided by hypotenuse, so the hypotenuse length must be calculated first.

3

Apply the cosine ratio definition

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

and simplify

\cos(\theta) = \frac{15}{10\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{3\sqrt{3}}{6} = \frac{\sqrt{3}}{2}

.

Rationalizing the denominator yields the simplified exact ratio.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:1m 15s

Question 94Question →

For any real number $x$ such that $0 < x < 3$ , which of the following expressions is equivalent to $\sin\left(\arccos\left(\frac{x}{3}\right)\right)$ ?

\frac{\sqrt{9 - x^2}}{3}

\frac{3 - x}{3}

\frac{\sqrt{x^2 - 9}}{3}

\frac{x}{\sqrt{9 - x^2}}

\frac{3}{x}

Show answer & explanation

Answer: $\frac{\sqrt{9 - x^2}}{3}$

Answer

\frac{\sqrt{9 - x^2}}{3}

By letting

\theta = \arccos\left(\frac{x}{3}\right)

, we construct a right triangle with an adjacent side of length

x

and a hypotenuse of length

3

. Applying the Pythagorean theorem yields an opposite side length of

\sqrt{3^2 - x^2} = \sqrt{9 - x^2}

. The sine of

\theta

is the ratio of the opposite side to the hypotenuse, which evaluates to

\frac{\sqrt{9 - x^2}}{3}

.

Step-by-Step Solution

1

Define an angle variable for the inverse trigonometric expression

Let

\theta = \arccos\left(\frac{x}{3}\right)

, so

\cos(\theta) = \frac{x}{3}

for

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

.

Setting the inverse trigonometric function to an angle allows setting up a right triangle relationship.

2

Set up side lengths of a right triangle using the definition of cosine

Adjacent side =

x

, Hypotenuse =

3

.

By definition,

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

.

3

Calculate the opposite side using the Pythagorean theorem

\text{opposite} = \sqrt{3^2 - x^2} = \sqrt{9 - x^2}.

In any right triangle,

\text{opposite}^2 + \text{adjacent}^2 = \text{hypotenuse}^2

.

4

Determine the sine of the angle

\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{9 - x^2}}{3}.

By definition,

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

.

Key Concept

Evaluating algebraic compositions of trigonometric and inverse trigonometric functions using right triangle geometry.

Estimated Time:1m 15s

Question 95Question →

In right triangle $PQR$ , the right angle is located at vertex $Q$ . The length of leg $PQ$ is $30$ centimeters. If $\sin(P) = \frac{8}{17}$ , what is the length, in centimeters, of leg $QR$ ?

16

\frac{240}{17}

18

34

\frac{225}{4}

Show answer & explanation

Answer: $16$

Answer

16 centimeters

In right triangle

PQR

,

\sin(P) = \frac{QR}{PR} = \frac{8}{17}

. The corresponding cosine ratio is

\cos(P) = \frac{PQ}{PR} = \frac{15}{17}

. Given

PQ = 30

, we solve

\frac{30}{PR} = \frac{15}{17}

to find

PR = 34

. Then, leg

QR = 34 \times \frac{8}{17} = 16

.

Step-by-Step Solution

1

Express the given sine ratio in terms of the triangle sides.

\sin(P) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{QR}{PR} = \frac{8}{17}

By definition of the sine function in a right triangle.

2

Find the cosine ratio

\cos(P)

using the Pythagorean identity or the 8-15-17 right triangle ratio.

\cos(P) = \sqrt{1 - \sin^2(P)} = \sqrt{1 - \frac{64}{289}} = \frac{15}{17}

Cosine represents the ratio of the adjacent side (

PQ

) to the hypotenuse (

PR

).

3

Calculate the hypotenuse

PR

using the known leg

PQ = 30

.

\frac{PQ}{PR} = \frac{15}{17} \implies \frac{30}{PR} = \frac{15}{17} \implies PR = 34

Setting the adjacent side ratio equal to

\cos(P)

solves for the hypotenuse length.

4

Calculate the opposite leg

QR

.

QR = PR \times \sin(P) = 34 \times \frac{8}{17} = 16

Multiplying the hypotenuse by

\sin(P)

gives the opposite leg length.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:1m 15s

Question 96Question →

What is the value, in degrees, of the expression $\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arccos\left(-\frac{1}{2}\right)$ ?

Show answer & explanation

Answer: 180

Answer

The value of the expression is 180 degrees.

To evaluate

\arcsin\left(\frac{\sqrt{3}}{2}\right) + \arccos\left(-\frac{1}{2}\right)

, find the principal value for each term. The principal range for

\arcsin

is

[-90^\circ, 90^\circ]

, so

\arcsin\left(\frac{\sqrt{3}}{2}\right) = 60^\circ

. The principal range for

\arccos

is

[0^\circ, 180^\circ]

, so for a negative argument, the result lies in Quadrant II, giving

\arccos\left(-\frac{1}{2}\right) = 120^\circ

. Adding these values together gives

60^\circ + 120^\circ = 180^\circ

.

Step-by-Step Solution

1

Find the principal angle for

\arcsin\left(\frac{\sqrt{3}}{2}\right)

in degrees.

\arcsin\left(\frac{\sqrt{3}}{2}\right) = 60^\circ

The inverse sine function yields outputs restricted to the range

[-90^\circ, 90^\circ]

. The angle in Quadrant I whose sine is

\frac{\sqrt{3}}{2}

is

60^\circ

.

2

Find the principal angle for

\arccos\left(-\frac{1}{2}\right)

in degrees.

\arccos\left(-\frac{1}{2}\right) = 120^\circ

The inverse cosine function yields outputs restricted to the range

[0^\circ, 180^\circ]

. For a negative input, the output must be in Quadrant II. The angle whose cosine is

-\frac{1}{2}

is

120^\circ

.

3

Sum the two evaluated angle measures.

60^\circ + 120^\circ = 180^\circ

Perform standard addition on the two principal values.

Key Concept

Evaluating inverse trigonometric functions within their standard principal value ranges.

Question 97Question →

What is the exact value of $\csc\left(\arctan\left(-\frac{7}{24}\right)\right)$ ?

-\frac{25}{7}

\frac{25}{7}

-\frac{25}{24}

-\frac{24}{7}

-\frac{7}{25}

Show answer & explanation

Answer: $-\frac{25}{7}$

Answer

-\frac{25}{7}

Let

\theta = \arctan\left(-\frac{7}{24}\right)

. By definition of the inverse tangent function's principal range

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

, a negative input produces an angle in Quadrant IV. In Quadrant IV, the opposite side is

-7

, the adjacent side is

24

, and the hypotenuse is

\sqrt{24^2 + (-7)^2} = 25

. Cosecant is the ratio of hypotenuse to opposite, giving

\frac{25}{-7} = -\frac{25}{7}

.

Step-by-Step Solution

1

Determine the quadrant for

\theta = \arctan\left(-\frac{7}{24}\right)

\theta

lies in Quadrant IV because the principal range of

\arctan(x)

is

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

and the input is negative.

Inverse tangent maps negative real numbers to angles in the interval

\left(-\frac{\pi}{2}, 0\right)

.

2

Set up a right triangle ratio for

\tan(\theta) = -\frac{7}{24}

Opposite side =

-7

, adjacent side =

24

, hypotenuse =

\sqrt{24^2 + (-7)^2} = 25

.

Tangent is the ratio of opposite to adjacent sides, and the hypotenuse is determined using the Pythagorean theorem.

3

Calculate

\csc(\theta)

\csc(\theta) = \frac{\text{hypotenuse}}{\text{opposite}} = \frac{25}{-7} = -\frac{25}{7}

.

Cosecant is defined as the reciprocal of sine, which equals hypotenuse divided by opposite side.

Key Concept

Evaluating trigonometric compositions involving inverse trigonometric functions using right triangle geometry and principal angle ranges.

Question 98Question →

A kite string of length $50$ meters is attached to a stake anchored in level ground at point $K$ . The kite is flying at point $H$ , directly above a landmark $L$ on the ground, forming right triangle $KLH$ with the right angle at $L$ . If the angle of elevation from the stake to the kite is $\theta$ , such that $\cos(\theta) = \frac{24}{25}$ , what is the vertical height, in meters, of the kite above the ground?

7

14

24

48

26

Show answer & explanation

Answer: $14$

Answer

The vertical height of the kite above the ground is

14

meters.

The vertical height corresponds to the leg opposite to angle

\theta

. Using

\cos(\theta) = \frac{24}{25}

, the adjacent side is

48

meters. Applying the Pythagorean theorem

LH = \sqrt{50^2 - 48^2} = 14

meters gives the correct vertical height.

Step-by-Step Solution

1

Identify the given trigonometric ratio and side length

Hypotenuse

KH = 50

meters and

\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{KL}{50} = \frac{24}{25}

.

Cosine relates the adjacent side (ground distance) to the hypotenuse (string length).

2

Calculate the horizontal ground distance

KL

KL = 50 \times \frac{24}{25} = 48

meters.

Multiply the hypotenuse length by the cosine ratio.

3

Determine the sine ratio or use the Pythagorean theorem to find the vertical height

LH

Since

\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(\frac{24}{25}\right)^2} = \frac{7}{25}

, the height

LH = 50 \times \sin(\theta) = 50 \times \frac{7}{25} = 14

meters.

The sine ratio relates the opposite vertical height to the hypotenuse.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Triples

Estimated Time:1m 0s

Question 99Question →

If $\theta = \arcsin\left(-\frac{3}{5}\right)$ , what is the exact decimal value of $\cos(2\theta)$ ?

Show answer & explanation

Answer: 0.28

Answer

The exact decimal value of

\cos(2\theta)

is

0.28

.

Given

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

, the sine of

\theta

is

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

. Using the double-angle formula for cosine,

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

, we substitute

\sin(\theta)

to get

\cos(2\theta) = 1 - 2\left(-\frac{3}{5}\right)^2 = 1 - \frac{18}{25} = \frac{7}{25} = 0.28

.

Step-by-Step Solution

1

Identify the value of

\sin(\theta)

from the inverse trigonometric expression

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

By definition of the inverse sine function, if

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

, then

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

where

-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}

.

2

Select the double-angle identity for cosine that uses sine

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

This form of the double-angle identity allows direct calculation without needing to calculate

\cos(\theta)

first.

3

Substitute

\sin(\theta)

and evaluate the expression

\cos(2\theta) = 0.28

Substituting

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

gives

1 - 2\left(-\frac{3}{5}\right)^2 = 1 - 2\left(\frac{9}{25}\right) = 1 - \frac{18}{25} = \frac{7}{25} = 0.28

.

Key Concept

Evaluating Trigonometric Functions of Inverse Trigonometric Expressions using Double-Angle Identities

Alternative Method

Alternatively, place

\theta

in Quadrant IV (since

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

for a negative inverse sine input). The adjacent side is

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

, so

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

. Then apply the alternative double-angle identity

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

.

Estimated Time:1m 15s

Question 100Question →

In right triangle $LMN$ , the right angle is at vertex $M$ . The hypotenuse $LN$ has a length of $25$ units. If $\cos(L) = \frac{7}{25}$ , what is the value of $\tan(N)$ ?

\frac{7}{24}

\frac{24}{7}

\frac{7}{25}

\frac{24}{25}

\frac{25}{7}

Show answer & explanation

Answer: $\frac{7}{24}$

Answer

\frac{7}{24}

The ratio

\cos(L) = \frac{7}{25}

means the leg adjacent to angle

L

(

LM

) is

7

units and the hypotenuse (

LN

) is

25

units. Using the Pythagorean theorem (

7^2 + MN^2 = 25^2

), the remaining leg

MN

is

24

units. For angle

N

, the opposite side is

LM = 7

and the adjacent side is

MN = 24

. Thus,

\tan(N) = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24}

.

Step-by-Step Solution

1

Use the definition of cosine for angle

L

to identify the length of leg

LM

.

Since

\cos(L) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{LM}{25} = \frac{7}{25}

, the length of leg

LM

is

7

units.

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

2

Calculate the length of the remaining leg

MN

using the Pythagorean theorem.

MN = \sqrt{LN^2 - LM^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24

units.

In a right triangle, the square of the hypotenuse equals the sum of the squares of the legs.

3

Set up the tangent ratio for angle

N

.

\tan(N) = \frac{\text{opposite side to } N}{\text{adjacent side to } N} = \frac{LM}{MN} = \frac{7}{24}

.

Tangent is defined as the ratio of the side opposite the angle to the side adjacent to the angle.

Key Concept

Right Triangle Trigonometric Ratios (SOHCAHTOA)

Estimated Time:1m 0s

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