Law of Sines and Law of Cosines Practice Questions — ACT

Question 1Question →

In triangle $ABC$ , the length of side $a$ is $5$ centimeters, the length of side $b$ is $8$ centimeters, and the measure of angle $C$ is $60^\circ$ . What is the length, in centimeters, of side $c$ ?

Show answer & explanation

Answer: 7

Answer

The length of side

c

is

7

centimeters.

Applying the Law of Cosines directly to the given Side-Angle-Side (SAS) triangle yields

c^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ = 25 + 64 - 40 = 49

, which gives

c = 7

.

Step-by-Step Solution

1

Identify the given values and the appropriate formula.

We are given two sides,

a = 5

and

b = 8

, and the included angle

C = 60^\circ

. To find the opposite side

c

, we use the Law of Cosines:

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

.

The Law of Cosines relates three sides of a triangle to the cosine of one of its angles, which is applicable for Side-Angle-Side (SAS) configurations.

2

Substitute the known values into the Law of Cosines equation.

c^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ

Plugging the given values into the formula allows us to solve for the unknown side

c

.

3

Evaluate the trigonometric and arithmetic terms.

Since

\cos 60^\circ = 0.5

, we get:

c^2 = 25 + 64 - 80(0.5)

c^2 = 89 - 40

c^2 = 49

Simplifying the expression step-by-step leads to the value of

c^2

.

4

Take the square root of both sides to find the side length.

c = \sqrt{49} = 7

Since side lengths must be positive, the square root of

49

gives the exact length of side

c

.

Key Concept

Law of Cosines

Question 2Question →

In triangle $PQR$ , the measure of angle $P$ is $30^\circ$ , the measure of angle $Q$ is $45^\circ$ , and the length of side $QR$ is $10$ units. Which of the following expressions represents the length, in units, of side $PR$ ?

10\sin(30^\circ)\sin(45^\circ)

\frac{10\sin(30^\circ)}{\sin(45^\circ)}

\frac{10\cos(45^\circ)}{\cos(30^\circ)}

\frac{10\sin(45^\circ)}{\sin(30^\circ)}

\frac{10\cos(30^\circ)}{\sin(45^\circ)}

Show answer & explanation

Answer: $\frac{10\sin(45^\circ)}{\sin(30^\circ)}$

Answer

The length of side

PR

is represented by the expression

\frac{10\sin(45^\circ)}{\sin(30^\circ)}

.

The correct answer is derived by setting up the Law of Sines proportion:

\frac{PR}{\sin(45^\circ)} = \frac{10}{\sin(30^\circ)}

. Multiplying both sides by

\sin(45^\circ)

yields

PR = \frac{10\sin(45^\circ)}{\sin(30^\circ)}

.

Step-by-Step Solution

1

Identify the known values and corresponding angle-side pairs in triangle

PQR

.

Angle

P = 30^\circ

is opposite to side

QR = 10

, and angle

Q = 45^\circ

is opposite to side

PR

.

This allows us to set up the appropriate trigonometric relationship to solve for the unknown side.

2

Apply the Law of Sines to relate the ratios of the side lengths to the sines of their opposite angles.

\frac{PR}{\sin(Q)} = \frac{QR}{\sin(P)}

, which becomes

\frac{PR}{\sin(45^\circ)} = \frac{10}{\sin(30^\circ)}

.

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

3

Isolate the variable representing the length of side

PR

.

PR = \frac{10\sin(45^\circ)}{\sin(30^\circ)}

.

Multiply both sides of the equation by

\sin(45^\circ)

to solve for

PR

.

Key Concept

The Law of Sines relates the side lengths of a triangle to the sines of its angles:

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

.

Estimated Time:1m 0s

Question 3Question →

A surveyor stands at point $A$ on flat ground and measures the angle of elevation to the top of a vertical tower, $T$ , at point $C$ to be $30^\circ$ . Another surveyor at point $B$ , which is $100$ meters away from $A$ on the same flat ground, measures $\angle CAB = 40^\circ$ and $\angle CBA = 65^\circ$ . Which of the following expressions represents the height, in meters, of the tower?

\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}

\frac{100 \sin(40^\circ) \tan(30^\circ)}{\sin(75^\circ)}

\frac{100 \sin(65^\circ) \sin(30^\circ)}{\sin(75^\circ)}

\frac{100 \sin(75^\circ) \tan(30^\circ)}{\sin(65^\circ)}

\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(105^\circ)}

Show answer & explanation

Answer: $\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}$

Answer

The correct expression is

\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}

The correct expression is derived by first applying the Law of Sines to find the length of the ground segment

AC

, and then using right-triangle trigonometry to determine the height of the tower. In triangle

ABC

, the third angle

\angle ACB

is

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

. The Law of Sines gives

\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)}

, which simplifies to

AC = \frac{100\sin(65^\circ)}{\sin(75^\circ)}

. Since the tower is vertical, triangle

ACT

is a right triangle with

\tan(30^\circ) = \frac{h}{AC}

. Substituting

AC

yields

h = \frac{100\sin(65^\circ)\tan(30^\circ)}{\sin(75^\circ)}

.

Step-by-Step Solution

1

Calculate the measure of the third angle in the ground triangle

\triangle ABC

.

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

The sum of angles in any triangle must equal

180^\circ

.

2

Apply the Law of Sines to find the distance from point

A

to the base of the tower at point

C

(

AC

).

\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)} \Rightarrow AC = \frac{100 \sin(65^\circ)}{\sin(75^\circ)}

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

3

Use the right-triangle trigonometric ratio for the vertical tower height

h

from point

A

.

\tan(30^\circ) = \frac{h}{AC} \Rightarrow h = AC \tan(30^\circ)

In right triangle

\triangle ACT

, the tangent of the angle of elevation is the ratio of the opposite side (height

h

) to the adjacent side (

AC

).

4

Substitute the expression for

AC

into the equation for

h

.

h = \frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}

Replacing

AC

with its equivalent algebraic expression yields the final height in terms of the given parameters.

Key Concept

Applying the Law of Sines to find a missing side length in a non-right triangle and then using right-triangle trigonometric ratios to solve a 3D geometry problem.

Estimated Time:2m 0s

Question 4Question →

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

Show answer & explanation

Answer: 45

Answer

The measure of the acute angle

ABC

is

45

degrees.

Applying the Law of Sines yields the relation

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

. Solving for

\sin(B)

yields

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

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

45^\circ

.

Step-by-Step Solution

1

Set up the Law of Sines relationship for triangle

ABC

using the side opposite angle

B

(

AC

) and the side opposite angle

A

(

BC

).

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

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

2

Substitute the given values into the equation:

AC = 12

,

BC = 6\sqrt{2}

, and

A = 30^\circ

.

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

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

B

.

3

Simplify the expression using the trigonometric value

\sin(30^\circ) = 0.5

and isolate

\sin(B)

.

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

Simplifying the fractions helps us identify the standard trigonometric value.

4

Find the acute angle

B

whose sine value is

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

.

B = 45^\circ

The inverse sine of

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

for an acute angle is

45^\circ

.

Key Concept

Law of Sines

Estimated Time:1m 0s

Question 5Question →

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

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

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

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

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

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

Show answer & explanation

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

Answer

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

The expression derived by applying the Law of Sines

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

correctly matches side

YZ

with its opposite angle

\angle X = 40^\circ

and side

XY = 14

with its opposite angle

\angle Z = 80^\circ

, yielding

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

.

Step-by-Step Solution

1

Identify the relevant law and relate the known sides and angles

Using the Law of Sines:

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

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

2

Substitute the given values into the formula

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

Side

XY = 14

is opposite

\angle Z = 80^\circ

, and side

YZ

is opposite

\angle X = 40^\circ

.

3

Solve for the unknown side

YZ

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

Multiply both sides of the equation by

\sin(40^\circ)

to isolate

YZ

.

Key Concept

Law of Sines

Question 6Question →

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

\frac{40}{9}

\frac{80}{9}

\frac{20}{9}

\frac{40}{3}

4\sqrt{3}

Show answer & explanation

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

Answer

\frac{40}{9}

meters

The total area of triangle

PQR

is equal to the sum of the areas of triangles

PQM

and

MQR

. Using the formula

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

, the equation

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

simplifies to

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

, which gives

x = \frac{40}{9}

.

Step-by-Step Solution

1

Express the area of the entire triangle

PQR

using the sine area formula.

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

square meters.

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

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

.

2

Express the sum of the areas of the two smaller triangles

\triangle PQM

and

\triangle MQR

created by the angle bisector

QM = x

.

Since

QM

bisects

\angle PQR = 120^\circ

,

\angle PQM = 60^\circ

and

\angle MQR = 60^\circ

.

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

.

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

.

An angle bisector divides the total angle into two equal

60^\circ

angles.

3

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

x

.

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

.

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

Key Concept

Area of Triangles using Sine and Angle Bisector Properties

Question 7Question →

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

Show answer & explanation

Answer: 11

Answer

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

Applying the Law of Cosines

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

with

a=8

,

b=15

, and

c=13

yields

64 = 225 + 169 - 390 \cos A

, which simplifies to

390 \cos A = 330

, giving

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

. In right triangle

ABD

,

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

, so

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

meters.

Step-by-Step Solution

1

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

cos A = 11/13

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

2

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

AD = 11 meters

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

Key Concept

Law of Cosines and Right Triangle Trigonometry

Estimated Time:2m 0s

Question 8Question →

In quadrilateral $ABCD$ , diagonal $AC$ divides the figure into two triangles, $\triangle ABC$ and $\triangle ACD$ . It is given that $AB = 6$ , $BC = 10$ , $\angle ABC = 120^\circ$ , $\angle CAD = 45^\circ$ , and $\angle ADC = 60^\circ$ . What is the length of side $CD$ ?

\frac{14\sqrt{6}}{3}

14

7\sqrt{3}

\frac{14\sqrt{2}}{3}

\sqrt{76}

Show answer & explanation

Answer: $\frac{14\sqrt{6}}{3}$

Answer

The length of side

CD

is

\frac{14\sqrt{6}}{3}

.

First, apply the Law of Cosines to

\triangle ABC

to find the length of diagonal

AC

:

AC^2 = 6^2 + 10^2 - 2(6)(10)\cos(120^\circ) = 36 + 100 - 120(-0.5) = 196

, which yields

AC = 14

. Next, use the Law of Sines in

\triangle ACD

:

\frac{CD}{\sin(45^\circ)} = \frac{14}{\sin(60^\circ)}

. Solving for

CD

gives

CD = 14 \cdot \frac{\sqrt{2}/2}{\sqrt{3}/2} = \frac{14\sqrt{2}}{\sqrt{3}} = \frac{14\sqrt{6}}{3}

.

Step-by-Step Solution

1

Apply the Law of Cosines in

\triangle ABC

to calculate the length of diagonal

AC

.

AC^2 = 6^2 + 10^2 - 2(6)(10)\cos(120^\circ) = 36 + 100 - 120\left(-\frac{1}{2}\right) = 136 + 60 = 196

, so

AC = 14

.

Two side lengths and the included angle of

\triangle ABC

are known.

2

Apply the Law of Sines in

\triangle ACD

to set up a proportion for side

CD

.

\frac{CD}{\sin(\angle CAD)} = \frac{AC}{\sin(\angle ADC)} \implies \frac{CD}{\sin(45^\circ)} = \frac{14}{\sin(60^\circ)}

.

The Law of Sines relates opposite sides and angles in

\triangle ACD

.

3

Solve for

CD

and rationalize the denominator.

CD = 14 \cdot \frac{\sin(45^\circ)}{\sin(60^\circ)} = 14 \cdot \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = \frac{14\sqrt{2}}{\sqrt{3}} = \frac{14\sqrt{6}}{3}

.

Evaluating the exact trigonometric values yields the final simplified length.

Key Concept

Law of Sines and Law of Cosines in Composite Triangles

Estimated Time:2m 0s

Question 9Question →

A telecommunications company connects two remote cell towers, Tower X and Tower Y, to a central relay station R. The cable path from Relay Station R to Tower X is $7$ kilometers long, and the cable path from Relay Station R to Tower Y is $8$ kilometers long. The angle formed between the two cables at Relay Station R, $\angle XRY$ , measures $120^\circ$ . A straight wireless backup link is established directly between Tower X and Tower Y. What is the distance, in kilometers, of this direct wireless link?

Show answer & explanation

Answer: 13

Answer

The distance of the direct wireless link between Tower X and Tower Y is 13 kilometers.

Using the Law of Cosines with two side lengths of 7 km and 8 km and an included angle of 120° gives

XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ) = 49 + 64 - 112(-0.5) = 169

. Taking the square root gives 13 km.

Step-by-Step Solution

1

Identify known components of the triangle formed by the relay station and towers

Side

RX = 7\text{ km}

, side

RY = 8\text{ km}

, and included angle

\angle R = 120^\circ

Two sides and the included angle (SAS) are known, indicating that the Law of Cosines must be used to find the third side.

2

Apply the Law of Cosines formula for side

XY

XY^2 = 7^2 + 8^2 - 2(7)(8)\cos(120^\circ)

The Law of Cosines relates three sides of a triangle to the cosine of one of its angles:

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

.

3

Calculate the numeric value of

XY^2

XY^2 = 49 + 64 + 56 = 169

Since

\cos(120^\circ) = -0.5

, the term

-2(56)(-0.5)

evaluates to

+56

.

4

Take the principal square root of

169

XY = 13\text{ km}

Distance must be a positive length.

Key Concept

Law of Cosines (Side-Angle-Side configuration)

Question 10Question →

In $\triangle ABC$ , the length of side $a$ (opposite angle $A$ ) is $9$ inches, the length of side $b$ (opposite angle $B$ ) is $12$ inches, and the measure of angle $B$ is $45^\circ$ . What is the exact value of $\sin A$ ?

\frac{3\sqrt{2}}{8}

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

\frac{3\sqrt{2}}{4}

\frac{3}{8}

\frac{4\sqrt{2}}{3}

Show answer & explanation

Answer: $\frac{3\sqrt{2}}{8}$

Answer

The exact value of

\sin A

is

\frac{3\sqrt{2}}{8}

.

According to the Law of Sines,

\frac{\sin A}{a} = \frac{\sin B}{b}

. Substituting

a = 9

,

b = 12

, and

B = 45^\circ

yields

\frac{\sin A}{9} = \frac{\sin 45^\circ}{12}

. Multiplying both sides by

9

and substituting

\sin 45^\circ = \frac{\sqrt{2}}{2}

gives

\sin A = \frac{9 \cdot \frac{\sqrt{2}}{2}}{12} = \frac{9\sqrt{2}}{24} = \frac{3\sqrt{2}}{8}

.

Step-by-Step Solution

1

State the Law of Sines for the given triangle components.

\frac{\sin A}{a} = \frac{\sin B}{b}

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

2

Substitute the known values (

a = 9

,

b = 12

, and

B = 45^\circ

) into the formula.

\frac{\sin A}{9} = \frac{\sin 45^\circ}{12}

Plugging in the given numbers isolates the unknown quantity

\sin A

.

3

Solve for

\sin A

and substitute the exact value of

\sin 45^\circ = \frac{\sqrt{2}}{2}

.

\sin A = \frac{9 \cdot \sin 45^\circ}{12} = \frac{9 \cdot \frac{\sqrt{2}}{2}}{12} = \frac{9\sqrt{2}}{24} = \frac{3\sqrt{2}}{8}

Simplifying the fraction gives the exact trigonometric ratio.

Key Concept

Law of Sines

Question 11Question →

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 12Question →

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 13Question →

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 14Question →

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 15Question →

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 16Question →

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 17Question →

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.

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