A landscape architect is designing a triangular courtyard garden. Two adjacent edges of th… — ACT

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?

Answer: 13 meters

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

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.

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

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

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