A plane is located at C on the diagram. There are two towers located at A and B. The distance between the towers is 7,600 feet, and the angles of elevation are given.

a. Find BC, the distance from Tower 2 to the plane, to the nearest foot.
b. Find CD, the height of the plane from the ground, to the nearest foot.

Tower 1:
Base angle = 16°

Tower 2:
Base angle = 24°

Base angle (for both towers) = 7600 ft.

1 answer

To find BC, we can use the Law of Sines.

We know that angle A is 16°, angle B is 24°, and the side AB is 7600 feet. Since the sum of angles in a triangle is 180°, angle C is 180 - 16 - 24 = 140°.

Now, we can use the Law of Sines to find BC:

sin(A) / BC = sin(B) / AC

sin(16°) / BC = sin(24°) / 7600

BC = 7600 * sin(16°) / sin(24°) ≈ 4772 feet

a. Therefore, the distance from Tower 2 to the plane (BC) is approximately 4772 feet.

Next, we can use the angle of elevation from Tower 2 (24°) to find the height of the plane above the ground (CD). We can do this by using the sine function:

sin(24°) = CD / BC

CD = BC * sin(24°) ≈ 4772 * sin(24°) ≈ 1918 feet

b. Therefore, the height of the plane from the ground (CD) is approximately 1918 feet.