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
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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)
1 answer