Right Triangle Trigonometry (SOHCAHTOA) Practice Questions — ACT

Question 1Question →

In right triangle $PQR$ , the right angle is at vertex $Q$ . If the side lengths are $PQ = 15$ and $QR = 8$ , what is the value of $\tan(P)$ ?

\frac{15}{8}

\frac{15}{17}

\frac{8}{15}

\frac{8}{17}

\frac{17}{8}

Show answer & explanation

Answer: $\frac{8}{15}$

Answer

The tangent of angle

P

is

\frac{8}{15}

.

The tangent of angle

P

is the ratio of the opposite side (

QR = 8

) to the adjacent side (

PQ = 15

). This gives the value

\frac{8}{15}

.

Step-by-Step Solution

1

Identify the reference angle and the sides of the right triangle relative to it.

The reference angle is

P

. The side opposite to angle

P

is

QR

with a length of

8

. The side adjacent to angle

P

is

PQ

with a length of

15

. The hypotenuse is

PR

.

To calculate a trigonometric ratio, we must first determine which sides are opposite, adjacent, and the hypotenuse relative to the target angle.

2

Recall the definition of the tangent ratio in a right triangle.

The tangent of an angle is defined as the ratio of the length of the opposite side to the length of the adjacent side:

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

.

The question asks for the tangent of angle

P

.

3

Substitute the identified side lengths into the tangent formula.

\tan(P) = \frac{QR}{PQ} = \frac{8}{15}

.

Plugging the lengths of the opposite side (

8

) and the adjacent side (

15

) into the tangent ratio yields the final value.

Key Concept

Right triangle trigonometry ratios (SOHCAHTOA), specifically the tangent ratio definition.

Estimated Time:45s

Question 2Question →

A $10$ -foot ladder leans against a vertical wall. The base of the ladder is $6$ feet away from the bottom of the wall, and the top of the ladder reaches a height of $8$ feet up the wall. If $\theta$ is the angle formed between the ladder and the ground, what is the value of $\cos(\theta)$ ?

\frac{3}{4}

\frac{3}{5}

\frac{4}{5}

\frac{4}{3}

\frac{5}{3}

Show answer & explanation

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

Answer

The correct answer is

\frac{3}{5}

, which represents the ratio of the adjacent side (

6

feet) to the hypotenuse (

10

feet) for the angle

\theta

formed between the ladder and the ground.

The correct answer is the value

\frac{3}{5}

. The angle

\theta

is formed between the ladder (hypotenuse) and the ground (adjacent side). The cosine of an angle in a right triangle is defined as the ratio of the length of the adjacent side to the length of the hypotenuse. Here, the adjacent side is

6

feet and the hypotenuse is

10

feet. Therefore,

\cos(\theta) = \frac{6}{10}

, which simplifies to

\frac{3}{5}

.

Step-by-Step Solution

1

Identify the parts of the right triangle relative to the angle

\theta

formed between the ladder and the ground.

The hypotenuse is the length of the ladder (

10

feet). The side adjacent to

\theta

is the distance along the ground from the wall to the base of the ladder (

6

feet). The side opposite to

\theta

is the height up the wall (

8

feet).

To apply trigonometric ratios, we must first map the given dimensions of the scenario to the sides of a right triangle relative to the reference angle.

2

Recall the definition of the cosine ratio in a right triangle.

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

We need to identify the mathematical definition of cosine to set up the calculation.

3

Substitute the identified side lengths into the cosine ratio and simplify the fraction.

\cos(\theta) = \frac{6}{10} = \frac{3}{5}

Substituting the values of

6

feet for the adjacent side and

10

feet for the hypotenuse gives

\frac{6}{10}

, which simplifies to

\frac{3}{5}

when both the numerator and denominator are divided by their greatest common divisor,

2

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Question 3Question →

In right triangle $ABC$ , the right angle is at vertex $B$ , $AB = 12$ , and $BC = 5$ . A line segment $BD$ is drawn perpendicular to the hypotenuse $AC$ such that $D$ lies on $AC$ . From point $D$ , a perpendicular line segment $DE$ is drawn to side $AB$ , where $E$ lies on $AB$ . What is the length of segment $DE$ ?

\frac{300}{169}

\frac{60}{13}

\frac{720}{169}

\frac{144}{13}

\frac{1728}{169}

Show answer & explanation

Answer: $\frac{720}{169}$

Answer

\frac{720}{169}

The correct answer is

\frac{720}{169}

. First, find the hypotenuse of the right triangle

ABC

using the Pythagorean theorem:

AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = 13

. In right triangle

ABC

, the sine of angle

A

is

\sin(\angle A) = \frac{BC}{AC} = \frac{5}{13}

. Next, in right triangle

ABD

(which has a right angle at

D

), the length of the altitude

BD

can be found using the sine of angle

A

:

BD = AB \sin(\angle A) = 12 \cdot \frac{5}{13} = \frac{60}{13}

. In right triangle

BDE

(which has a right angle at

E

), the angle

\angle BDE

is equal to

\angle A

because both are complementary to

\angle ABD

(or

\angle EBD

). Therefore, the adjacent side

DE

is found using the cosine of angle

\angle BDE

:

DE = BD \cos(\angle BDE) = BD \cos(\angle A) = \frac{60}{13} \cdot \frac{12}{13} = \frac{720}{169}

.

Step-by-Step Solution

1

Find the hypotenuse

AC

of the right triangle

ABC

using the Pythagorean theorem.

AC = 13

The hypotenuse is needed to find the trigonometric ratios of angle

A

.

2

Determine

\sin(\angle A)

and

\cos(\angle A)

from right triangle

ABC

.

\sin(\angle A) = \frac{5}{13}

and

\cos(\angle A) = \frac{12}{13}

These trigonometric ratios are needed for the calculations in the nested right triangles.

3

Find the length of altitude

BD

in right triangle

ABD

using

\sin(\angle A)

.

BD = \frac{60}{13}

BD

serves as the hypotenuse for the next right triangle

BDE

.

4

Identify that

\angle BDE = \angle A

and calculate the length of

DE

in right triangle

BDE

.

DE = \frac{720}{169}

Since

\angle BDE = \angle A

,

DE = BD \cos(\angle BDE) = BD \cos(\angle A) = \frac{60}{13} \cdot \frac{12}{13} = \frac{720}{169}

.

Key Concept

Applying SOHCAHTOA (sine and cosine definitions) sequentially across nested right triangles by identifying equal angles.

Estimated Time:2m 0s

Question 4Question →

A coordinate grid contains a right triangle, $XYZ$ , with the right angle located at vertex $Z$ . The horizontal leg $XZ$ has a length of $24$ units, and the vertical leg $YZ$ has a length of $7$ units. What is the value of $\cos(X)$ ?

\frac{7}{25}

\frac{24}{25}

\frac{7}{24}

\frac{24}{7}

\frac{25}{24}

Show answer & explanation

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

Answer

The correct answer is the option representing

\frac{24}{25}

.

The cosine of an angle in a right triangle is the ratio of the length of the adjacent leg to the length of the hypotenuse. The hypotenuse of the triangle is calculated as

25

units using the Pythagorean theorem. Relative to angle

X

, the adjacent leg is

XZ = 24

and the hypotenuse is

XY = 25

. Therefore,

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

.

Step-by-Step Solution

1

Use the Pythagorean theorem to calculate the hypotenuse

XY

.

XY = \sqrt{XZ^2 + YZ^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25

.

To calculate any primary trigonometric ratio, we need the lengths of the relevant sides, including the hypotenuse.

2

Identify the adjacent side relative to angle

X

.

The leg adjacent to angle

X

is

XZ = 24

.

Angle

X

is formed by the hypotenuse

XY

and the adjacent leg

XZ

.

3

Apply the definition of cosine (adjacent over hypotenuse).

\cos(X) = \frac{XZ}{XY} = \frac{24}{25}

.

By SOHCAHTOA, cosine is the ratio of the adjacent side length to the hypotenuse length.

Key Concept

Using SOHCAHTOA definitions to find trigonometric ratios in a right triangle.

Estimated Time:1m 0s

Question 5Question →

A straight skateboard ramp has a vertical height of $9$ feet and a horizontal length of $12$ feet along the ground. The vertical height and horizontal length meet at a right angle. What is the sine of the angle that the ramp makes with the ground?

\frac{3}{4}

\frac{4}{5}

\frac{3}{5}

\frac{4}{3}

\frac{5}{3}

Show answer & explanation

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

Answer

The value of the sine of the angle is

\frac{3}{5}

.

The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. The opposite side is the vertical height of

9

feet. Using the Pythagorean theorem, the hypotenuse (the length of the ramp) is calculated as

\sqrt{9^2 + 12^2} = 15

feet. Therefore, the sine of the angle is

\frac{9}{15}

, which simplifies to

\frac{3}{5}

.

Step-by-Step Solution

1

Identify the lengths of the legs of the right triangle and calculate the length of the hypotenuse using the Pythagorean theorem.

The hypotenuse (ramp length) is

15

feet.

The vertical height (

9

feet) and horizontal length (

12

feet) form the perpendicular legs of a right triangle. The length of the ramp is the hypotenuse:

\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15

feet.

2

Identify the opposite side and hypotenuse relative to the angle the ramp makes with the ground, then apply the sine ratio definition.

The sine of the angle is

\frac{3}{5}

.

The angle the ramp makes with the ground is at the bottom vertex. The side opposite this angle is the vertical height (

9

feet), and the hypotenuse is the ramp length (

15

feet). The sine ratio is

\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{15}

, which simplifies to

\frac{3}{5}

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA)

Estimated Time:1m 0s

Question 6Question →

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 7Question →

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 8Question →

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 9Question →

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 10Question →

In right triangle $XYZ$ , the right angle is at vertex $Y$ , and line segment $YW$ is an altitude perpendicular to hypotenuse $XZ$ at point $W$ . If the length of side $XY$ is $15$ units and $\cos(X) = \frac{4}{5}$ , what is the length of line segment $ZW$ ?

9

12

\frac{27}{4}

\frac{45}{4}

16

Show answer & explanation

Answer: $\frac{27}{4}$

Answer

The length of line segment

ZW

is

\frac{27}{4}

units.

In right triangle

XYZ

,

\cos(X) = \frac{XY}{XZ}

. Given

XY = 15

and

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

, we solve for hypotenuse

XZ = \frac{75}{4}

. In right triangle

XYW

,

\cos(X) = \frac{XW}{XY} = \frac{XW}{15}

, yielding

XW = 12

. Subtracting

XW

from total hypotenuse

XZ

yields

ZW = \frac{75}{4} - 12 = \frac{27}{4}

.

Step-by-Step Solution

1

Find the length of hypotenuse

XZ

using

\cos(X)

in

\triangle XYZ

.

XZ = \frac{75}{4}

In

\triangle XYZ

,

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

. Substituting

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

and

XY = 15

gives

\frac{4}{5} = \frac{15}{XZ}

, so

XZ = \frac{15 \times 5}{4} = \frac{75}{4}

.

2

Find the length of segment

XW

using

\cos(X)

in right triangle

XYW

.

XW = 12

In right triangle

XYW

(with right angle at

W

),

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

. Substituting

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

and

XY = 15

gives

\frac{4}{5} = \frac{XW}{15}

, so

XW = 12

.

3

Calculate segment

ZW

by subtracting

XW

from total hypotenuse

XZ

.

ZW = \frac{27}{4}

Since point

W

lies on segment

XZ

,

ZW = XZ - XW = \frac{75}{4} - 12 = \frac{75}{4} - \frac{48}{4} = \frac{27}{4}

.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) in Nested Right Triangles

Question 11Question →

In right triangle $ABC$ , the right angle is located at vertex $C$ . Point $M$ is the midpoint of leg $BC$ . The length of leg $AC$ is $12$ units, and $\tan(\angle MAC) = \frac{1}{3}$ . What is the value of $\sin(\angle BAC)$ ?

\frac{2\sqrt{13}}{13}

\frac{3\sqrt{13}}{13}

\frac{2}{3}

\frac{\sqrt{10}}{10}

\frac{2\sqrt{5}}{5}

Show answer & explanation

Answer: $\frac{2\sqrt{13}}{13}$

Answer

The value of

\sin(\angle BAC)

is

\frac{2\sqrt{13}}{13}

.

In right triangle

ACM

,

\tan(\angle MAC) = \frac{MC}{AC} = \frac{1}{3}

. Since

AC = 12

, we find

MC = 4

. Because

M

is the midpoint of side

BC

,

BC = 2 \times 4 = 8

. In right triangle

ABC

, the hypotenuse is

AB = \sqrt{12^2 + 8^2} = \sqrt{208} = 4\sqrt{13}

. The sine of angle

BAC

is defined as

\frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{8}{4\sqrt{13}} = \frac{2\sqrt{13}}{13}

.

Step-by-Step Solution

1

Find the length of segment

MC

using right triangle

ACM

.

MC = 4

In right triangle

ACM

with right angle at

C

,

\tan(\angle MAC) = \frac{\text{opposite}}{\text{adjacent}} = \frac{MC}{AC}

. Given

\tan(\angle MAC) = \frac{1}{3}

and

AC = 12

,

\frac{MC}{12} = \frac{1}{3} \implies MC = 4

.

2

Determine the length of side

BC

.

BC = 8

Since

M

is the midpoint of leg

BC

,

BC = 2 \times MC = 2 \times 4 = 8

.

3

Calculate hypotenuse

AB

of right triangle

ABC

.

AB = 4\sqrt{13}

By the Pythagorean theorem in

\triangle ABC

:

AB = \sqrt{AC^2 + BC^2} = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}

.

4

Calculate

\sin(\angle BAC)

.

\sin(\angle BAC) = \frac{2\sqrt{13}}{13}

In right triangle

ABC

,

\sin(\angle BAC) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{8}{4\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}

.

Key Concept

Applying SOHCAHTOA definitions and the Pythagorean theorem in multi-step right triangle geometry.

Estimated Time:2m 0s

Question 12Question →

In right triangle $PQR$ , the right angle is at vertex $Q$ . Point $S$ lies on leg $PQ$ such that the length of $PS$ is $7$ units and the length of $SQ$ is $5$ units. If $\tan(\angle PRQ) = \frac{4}{3}$ , what is the value of $\cos(\angle SRQ)$ ?

\frac{5\sqrt{106}}{106}

\frac{9\sqrt{106}}{106}

\frac{9\sqrt{130}}{130}

\frac{16\sqrt{281}}{281}

\frac{5}{9}

Show answer & explanation

Answer: $\frac{9\sqrt{106}}{106}$

Answer

The value of

\cos(\angle SRQ)

is

\frac{9\sqrt{106}}{106}

.

The total length of leg

PQ

is

7 + 5 = 12

. Using the tangent definition in right triangle

PQR

,

\tan(\angle PRQ) = \frac{PQ}{QR} = \frac{12}{QR} = \frac{4}{3}

, which gives

QR = 9

. Next, considering right triangle

SQR

with legs

SQ = 5

and

QR = 9

, the hypotenuse

SR = \sqrt{5^2 + 9^2} = \sqrt{106}

. Finally,

\cos(\angle SRQ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{QR}{SR} = \frac{9}{\sqrt{106}} = \frac{9\sqrt{106}}{106}

.

Step-by-Step Solution

1

Find the length of leg

PQ

PQ = PS + SQ = 7 + 5 = 12

units

Point

S

lies on segment

PQ

, so the total length is the sum of its parts.

2

Calculate the length of leg

QR

using

\tan(\angle PRQ)

\tan(\angle PRQ) = \frac{PQ}{QR} \implies \frac{4}{3} = \frac{12}{QR} \implies QR = 9

units

In right triangle

PQR

, tangent is opposite side over adjacent side relative to

\angle PRQ

.

3

Find hypotenuse

SR

of right triangle

SQR

SR = \sqrt{SQ^2 + QR^2} = \sqrt{5^2 + 9^2} = \sqrt{25 + 81} = \sqrt{106}

units

Apply the Pythagorean theorem to right triangle

SQR

with right angle at

Q

.

4

Determine

\cos(\angle SRQ)

and rationalize the denominator

\cos(\angle SRQ) = \frac{QR}{SR} = \frac{9}{\sqrt{106}} = \frac{9\sqrt{106}}{106}

Cosine is defined as the ratio of adjacent side over hypotenuse in right triangle

SQR

.

Key Concept

Applying SOHCAHTOA and the Pythagorean Theorem in composite right triangle figures

Estimated Time:2m 0s

Question 13Question →

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 14Question →

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 15Question →

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 16Question →

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 17Question →

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 18Question →

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 19Question →

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 20Question →

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

Right Triangle Trigonometry (SOHCAHTOA)

Answer

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