Right Triangle Trigonometry (SOHCAHTOA) Practice Questions — ACT — Page 2

Question 21Question →

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

Question 22Question →

A diagonal support beam is installed to stabilize a wooden wall frame. The beam extends from the top-left corner of the frame to the bottom-right corner, forming a right triangle with the top horizontal beam and the right vertical post. The vertical post has a height of $15$ feet. If the angle $\theta$ between the diagonal support beam and the top horizontal beam satisfies $\tan(\theta) = \frac{3}{4}$ , what is the length, in feet, of the diagonal support beam?

11.25

20

25

35

18.75

Show answer & explanation

Answer: $25$

Answer

The length of the diagonal support beam is

25

feet.

In the right triangle formed by the wall frame and diagonal beam, angle

\theta

is between the diagonal beam (hypotenuse) and the top horizontal beam (adjacent side). The vertical post is opposite to angle

\theta

, with a length of

15

feet. Using

\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4}

, we substitute

15

for the opposite side to get

\frac{3}{4} = \frac{15}{\text{Adjacent}}

, which yields an adjacent side length of

20

feet. Using the Pythagorean theorem, the hypotenuse is

\sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25

feet.

Step-by-Step Solution

1

Identify the given trigonometric relationship and right triangle sides.

The vertical post is opposite to angle

\theta

, so

\text{Opposite} = 15

feet. The top horizontal beam is adjacent to

\theta

, and the diagonal support beam is the hypotenuse.

Understanding side positions relative to angle

\theta

is essential for setting up the tangent ratio correctly.

2

Use the definition of tangent to find the length of the adjacent side (horizontal beam).

\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} \implies \frac{3}{4} = \frac{15}{\text{Adjacent}} \implies \text{Adjacent} = \frac{15 \times 4}{3} = 20

feet.

Solving for the adjacent side gives the horizontal leg of the right triangle.

3

Apply the Pythagorean theorem to calculate the length of the diagonal support beam (hypotenuse).

\text{Hypotenuse} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25

feet.

The diagonal beam forms the hypotenuse of the right triangle with legs of length

15

feet and

20

feet.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and the Pythagorean Theorem

Estimated Time:1m 15s

Question 23Question →

A surveyor standing at point $P$ on level ground measures the angle of elevation to the top of a cliff, point $T$ , as $\theta$ . The ground distance from point $P$ to the vertical base of the cliff, point $B$ , is $120$ feet. If $\cos(\theta) = \frac{12}{13}$ , what is the vertical height of the cliff, in feet?

50

65

110

130

288

Show answer & explanation

Answer: 50

Answer

The vertical height of the cliff is 50 feet.

In right triangle

\triangle PBT

with the right angle at

B

, point

P

on the ground forms angle

\theta

. The adjacent side

PB = 120

feet. Given

\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}

, we solve for hypotenuse

PT = 130

feet. Using the Pythagorean theorem (

130^2 - 120^2 = 50^2

), the opposite side

TB

(the vertical height of the cliff) is

50

feet.

Step-by-Step Solution

1

Identify the given trigonometric ratio and sides of the right triangle

\triangle PBT

.

Angle

\theta

is at vertex

P

, the adjacent leg

PB = 120

feet, and

\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}

.

Cosine relates the adjacent side to the hypotenuse in a right triangle.

2

Calculate the length of the hypotenuse

PT

.

\frac{120}{PT} = \frac{12}{13} \implies PT = \frac{120 \times 13}{12} = 130

feet.

Solve the ratio for the hypotenuse

PT

.

3

Find the vertical height

TB

using the Pythagorean theorem or sine ratio.

TB = \sqrt{PT^2 - PB^2} = \sqrt{130^2 - 120^2} = \sqrt{16900 - 14400} = \sqrt{2500} = 50

feet.

In right triangle

\triangle PBT

,

TB^2 + PB^2 = PT^2

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Triples

Estimated Time:1m 15s

Question 24Question →

In right triangle $XYZ$ , the right angle is located at vertex $Y$ . The hypotenuse $XZ$ has a length of $39$ inches. If $\cos(X) = \frac{5}{13}$ , what is the length, in inches, of leg $YZ$ ?

15

24

36

42

65

Show answer & explanation

Answer: 36

Answer

36 inches

By SOH CAH TOA,

\cos(X) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{39}

. Setting

\frac{XY}{39} = \frac{5}{13}

gives

XY = 15

inches. Applying the Pythagorean theorem to find the remaining leg

YZ

gives

YZ = \sqrt{39^2 - 15^2} = \sqrt{1296} = 36

inches. Alternatively, recognizing the ratio

\cos(X) = \frac{5}{13}

implies

\sin(X) = \frac{12}{13}

for a 5-12-13 right triangle, so

YZ = 39 \times \frac{12}{13} = 36

inches.

Step-by-Step Solution

1

Use the definition of cosine (SOH CAH TOA) to find the length of leg

XY

, which is adjacent to angle

X

.

\cos(X) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{XY}{XZ} \implies \frac{5}{13} = \frac{XY}{39} \implies XY = 39 \times \frac{5}{13} = 15

inches.

Cosine is defined as the ratio of the adjacent side to the hypotenuse.

2

Apply the Pythagorean theorem (

XY^2 + YZ^2 = XZ^2

) to calculate the length of leg

YZ

.

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

inches.

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

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Theorem

Estimated Time:1m 0s

Question 25Question →

In right triangle $CDE$ , the right angle is located at vertex $D$ . The length of leg $CD$ is $24$ centimeters. If $\tan(E) = \frac{4}{3}$ , what is the value of $\sin(C)$ ?

\frac{3}{5}

\frac{4}{5}

\frac{3}{4}

\frac{4}{3}

\frac{5}{3}

Show answer & explanation

Answer: $\frac{3}{5}$

Answer

The value of

\sin(C)

is

\frac{3}{5}

.

To find

\sin(C)

, first express

\tan(E) = \frac{\text{opposite}}{\text{adjacent}} = \frac{CD}{DE} = \frac{24}{DE} = \frac{4}{3}

, which gives

DE = 18

. Next, compute the hypotenuse

CE = \sqrt{24^2 + 18^2} = \sqrt{900} = 30

. Finally, identify the side opposite to angle

C

, which is

DE = 18

. Thus,

\sin(C) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{18}{30} = \frac{3}{5}

. The option stating

\frac{3}{5}

is correct.

Step-by-Step Solution

1

Use the definition of tangent for angle

E

to find the length of side

DE

.

DE = 18\text{ cm}

Since

\tan(E) = \frac{\text{opposite}}{\text{adjacent}} = \frac{CD}{DE}

, substituting

CD = 24

yields

\frac{24}{DE} = \frac{4}{3}

, so

DE = \frac{24 \times 3}{4} = 18

.

2

Calculate the hypotenuse

CE

using the Pythagorean theorem.

CE = 30\text{ cm}

In right triangle

CDE

,

CE^2 = CD^2 + DE^2 = 24^2 + 18^2 = 576 + 324 = 900

, so

CE = \sqrt{900} = 30

.

3

Calculate

\sin(C)

using the SOHCAHTOA ratio for angle

C

.

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

Relative to angle

C

, the opposite side is

DE = 18

and the hypotenuse is

CE = 30

. Thus,

\sin(C) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{18}{30} = \frac{3}{5}

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:1m 15s

Question 26Question →

In right triangle $DEF$ , the right angle is located at vertex $E$ . The length of leg $DE$ is $32$ meters. If $\cos(D) = \frac{8}{17}$ , what is the length of leg $EF$ , in meters?

15

36

60

68

136

Show answer & explanation

Answer: 60

Answer

The length of leg

EF

is 60 meters.

Using the definition of cosine,

\cos(D) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{DE}{DF}

. Substituting

DE = 32

gives

\frac{32}{DF} = \frac{8}{17}

, so

DF = 68

. Using the Pythagorean theorem,

EF = \sqrt{68^2 - 32^2} = 60

meters.

Step-by-Step Solution

1

Set up the cosine ratio for angle D

\cos(D) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{DE}{DF} = \frac{8}{17}

By SOHCAHTOA, cosine of an acute angle in a right triangle is the ratio of the adjacent leg to the hypotenuse.

2

Calculate the length of hypotenuse DF

\frac{32}{DF} = \frac{8}{17} \implies DF = \frac{32 \times 17}{8} = 68\text{ meters}

Substitute the given value

DE = 32

into the ratio and solve for

DF

.

3

Calculate the length of leg EF using the Pythagorean theorem

EF = \sqrt{DF^2 - DE^2} = \sqrt{68^2 - 32^2} = \sqrt{4624 - 1024} = \sqrt{3600} = 60\text{ meters}

In right triangle

DEF

,

DE^2 + EF^2 = DF^2

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) & Pythagorean Theorem

Question 27Question →

A straight ramp connects a driveway to a loading dock that is $10$ feet above the ground. The ramp forms an angle $\theta$ with the flat ground such that $\tan(\theta) = \frac{5}{12}$ . What is the length, in feet, of the ramp?

14

24

26

2\sqrt{61}

30

Show answer & explanation

Answer: $26$

Answer

The length of the ramp is

26

feet.

In the right triangle formed by the ground, the loading dock, and the ramp, the dock height (

10

feet) is opposite angle

\theta

, and the ramp is the hypotenuse. Since

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}

, setting

\frac{10}{\text{adjacent}} = \frac{5}{12}

yields an adjacent side of

24

feet. Applying the Pythagorean theorem to find the hypotenuse gives

\sqrt{10^2 + 24^2} = \sqrt{676} = 26

feet.

Step-by-Step Solution

1

Identify the given information and trigonometric ratio.

The vertical leg opposite angle

\theta

is

10

feet. The ratio is

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{12}

.

Tangent is defined as the ratio of the side opposite the angle to the side adjacent to the angle in a right triangle.

2

Solve for the length of the adjacent leg (ground distance).

\frac{10}{\text{adjacent}} = \frac{5}{12} \implies \text{adjacent} = \frac{10 \times 12}{5} = 24

feet.

Cross-multiplying gives the length of the ground leg.

3

Calculate the length of the ramp (hypotenuse) using the Pythagorean theorem.

\text{ramp length} = \sqrt{10^2 + 24^2} = \sqrt{100 + 576} = \sqrt{676} = 26\text{ feet}.

The ramp forms the hypotenuse of the right triangle, so its length is

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

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Theorem

Estimated Time:1m 15s

Question 28Question →

A technician is installing a straight support beam for a solar panel array mounted on a flat roof. The beam forms a right triangle with the horizontal roof and a vertical panel frame. The vertical frame is $24$ inches tall, and the angle $\theta$ between the support beam and the horizontal roof satisfies $\sin(\theta) = \frac{12}{13}$ . What is the horizontal distance, in inches, along the roof from the bottom of the vertical frame to the anchor point of the support beam?

5

10

12

24

26

Show answer & explanation

Answer: 10

Answer

The horizontal distance along the roof from the bottom of the frame to the anchor point is 10 inches.

In a right-angled triangle formed by the vertical frame, horizontal roof, and diagonal support beam, the angle

\theta

is between the beam and the roof. The vertical frame (24 inches) is the side opposite to

\theta

. Using the sine definition,

\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13}

, so

\frac{24}{\text{hypotenuse}} = \frac{12}{13}

, which yields a hypotenuse length of 26 inches. Applying the Pythagorean theorem to find the horizontal adjacent leg gives

\sqrt{26^2 - 24^2} = \sqrt{100} = 10

inches.

Step-by-Step Solution

1

Identify the sides of the right triangle relative to the angle

\theta

.

The vertical frame of length 24 inches is opposite to

\theta

, the support beam is the hypotenuse, and the horizontal distance along the roof is adjacent to

\theta

.

SOHCAHTOA defines

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

.

2

Calculate the length of the hypotenuse using the sine ratio.

\frac{24}{\text{hypotenuse}} = \frac{12}{13} \implies \text{hypotenuse} = 24 \times \frac{13}{12} = 26

inches.

Setting the opposite side (24) over hypotenuse equal to

\frac{12}{13}

allows solving for the hypotenuse.

3

Calculate the horizontal adjacent side using the Pythagorean theorem.

\text{adjacent} = \sqrt{26^2 - 24^2} = \sqrt{676 - 576} = \sqrt{100} = 10

inches.

In a right triangle,

a^2 + b^2 = c^2

, so the unknown leg is

\sqrt{c^2 - b^2}

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:1m 15s

Question 29Question →

In right triangle $UVW$ , the right angle is located at vertex $V$ . The length of hypotenuse $UW$ is $85$ centimeters. If $\sin(U) = \frac{15}{17}$ , what is the length, in centimeters, of leg $UV$ ?

40

45

68

75

114

Show answer & explanation

Answer: $40$

Answer

The length of leg

UV

is

40

centimeters.

By definition,

\sin(U) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{VW}{UW}

. Given

\sin(U) = \frac{15}{17}

and

UW = 85

, the opposite leg

VW = 85 \times \frac{15}{17} = 75

cm. Using the Pythagorean theorem

UV^2 = UW^2 - VW^2 = 85^2 - 75^2 = 7225 - 5625 = 1600

, we get

UV = 40

cm. Alternatively, using

\cos(U) = \sqrt{1 - \sin^2(U)} = \frac{8}{17}

, the adjacent leg

UV = 85 \times \frac{8}{17} = 40

cm.

Step-by-Step Solution

1

Identify the relationship between

\sin(U)

and the sides of right triangle

UVW

.

\sin(U) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{VW}{UW} = \frac{15}{17}

.

By SOHCAHTOA, sine is the ratio of the opposite leg to the hypotenuse.

2

Calculate the length of the opposite leg

VW

.

VW = 85 \times \frac{15}{17} = 75

centimeters.

Multiply the hypotenuse length

UW = 85

by the sine ratio

\frac{15}{17}

.

3

Apply the Pythagorean theorem (

UV^2 + VW^2 = UW^2

) to solve for adjacent leg

UV

.

UV^2 + 75^2 = 85^2 \implies UV^2 + 5625 = 7225 \implies UV^2 = 1600 \implies UV = 40

centimeters.

The square of the hypotenuse equals the sum of the squares of the two legs.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Relationship

Estimated Time:1m 30s

Question 30Question →

An observer stands at point $P$ on horizontal ground and measures the angle of elevation to the top of a vertical tower, $T$ , as $\theta$ . The observer then walks a distance of $d$ meters directly toward the base of the tower to point $Q$ . From point $Q$ , the angle of elevation to the top of the tower is $2\theta$ . If $\cos(\theta) = \frac{4}{5}$ , what is the ratio of the height of the tower to the distance $d$ ?

\frac{3}{5}

\frac{3}{4}

\frac{12}{25}

\frac{24}{25}

\frac{7}{25}

Show answer & explanation

Answer: $\frac{24}{25}$

Answer

\frac{24}{25}

The correct answer of

\frac{24}{25}

is found by first identifying that the triangle formed by the tower's top and the two observer positions is isosceles. Since the exterior angle is

2\theta

and one interior angle is

\theta

, the other interior angle must also be

\theta

, making the side lengths opposite these angles equal (

d

). By dropping an altitude inside this isosceles triangle, we form two right triangles, allowing us to find the hypotenuse of the larger right triangle as

2d\cos(\theta) = 1.6d

. Applying the sine definition to the larger right triangle yields

\sin(\theta) = \frac{h}{1.6d}

. Since

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

, we have

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

. Solving for

\frac{h}{d}

gives

\frac{h}{d} = 1.6 \times \frac{3}{5} = \frac{24}{25}

.

Step-by-Step Solution

1

Analyze the angles in the triangle formed by the top of the tower

T

, the first position

P

, and the second position

Q

.

The exterior angle at vertex

Q

is

2\theta

, and the opposite interior angle at vertex

P

is

\theta

. Therefore, the third angle

\angle PTQ

is

2\theta - \theta = \theta

. Since two angles are equal, triangle

PTQ

is isosceles with

QT = PQ = d

.

This identifies the length of segment

QT

in terms of the walking distance

d

.

2

Drop a perpendicular altitude from

Q

to segment

PT

meeting at point

M

. Use right triangle trigonometry in the resulting right triangle

PMQ

.

In right triangle

PMQ

, the hypotenuse is

PQ = d

and the angle is

\theta

. Thus, the adjacent side is

PM = d \cos(\theta) = \frac{4}{5}d = 0.8d

. Because the altitude of an isosceles triangle bisects the base, the total length

PT = 2 \times PM = \frac{8}{5}d = 1.6d

.

This determines the length of the hypotenuse

PT

of the large right triangle

PRT

in terms of

d

.

3

Apply the sine ratio to the large right triangle

PRT

to find the ratio of the height

h

to the distance

d

.

In right triangle

PRT

, the angle at

P

is

\theta

, the opposite side is

h

(height of the tower), and the hypotenuse is

PT = 1.6d

. Therefore,

\sin(\theta) = \frac{h}{PT} = \frac{h}{1.6d}

. Since

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

, we have

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

. Substituting this gives

\frac{3}{5} = \frac{h}{1.6d} \implies \frac{h}{d} = 1.6 \times \frac{3}{5} = \frac{24}{25}

.

This solves for the ratio of the tower height to the distance

d

.

Key Concept

Applying SOHCAHTOA and geometric properties of triangles to solve multi-step trigonometry problems

Estimated Time:3m 0s

Right Triangle Trigonometry (SOHCAHTOA)

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