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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.
Right triangle A C D is shown with point B between A and D with a line segment connecting points B and C dividing the larger triangle into triangles A B C and C B D.
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.
(3 points)
First, let's assume the angle of elevation at Tower A is 43 degrees, and at Tower B is 34 degrees.
a. To find BC, the distance from Tower B to the plane, we will use the tangent function in triangle CBP where P is the projection point from B to the plane.
tan(34) = CD / BC
tan(34) = CD / 7600
CD = 7600 * tan(34)
CD ≈ 4790 feet
b. To find CD, the height of the plane from the ground, we will use the tangent function in triangle ACD.
tan(43) = CD / AC
tan(43) = CD / 7600
CD = 7600 * tan(43)
CD ≈ 5785 feet
Therefore,
a. BC ≈ 4790 feet
b. CD ≈ 5785 feet