Questions: 65. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in Figure 33. Find the distance of the plane from point A to the nearest tenth of a kilometer.

65. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be 32° and 56°, as shown in Figure 33. Find the distance of the plane from point A to the nearest tenth of a kilometer.

Transcript text: 65. A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 4.3 km apart, to be $32^{\circ}$ and $56^{\circ}$, as shown in Figure 33. Find the distance of the plane from point $A$ to the nearest tenth of a kilometer. Figure 33

Solution

Solution Steps

Step 1: Find the distance from the plane to point B

Let P be the location of the plane, and let h be the altitude of the plane above the highway. We are given that AB = 4.3 km. Angle PBA is 56°. In triangle PBA, we have tan(56°) = h / PB. Thus, PB = h / tan(56°).

Step 2: Find the distance from the plane to point A

Angle PAB is 32°. The distance between the mileposts A and B is 4.3 km. In triangle PAB, PA is the hypotenuse. In triangle PBA, PB is the adjacent side to the 56° angle. In triangle PAA', PA is the hypotenuse. In triangle PBA, PB is the adjacent side to the angle of 56°. We can use the tangent function to relate h to PA. tan(32°) = h/PA. So, PA = h/tan(32°). Also, PB = h/tan(56°). Since A, B are on the same line, PA = PB + AB.

Step 3: Solve for h and PA

We know that AB = 4.3 km.
PA = PB + AB h/tan(32°) = h/tan(56°) + 4.3 h(1/tan(32°) - 1/tan(56°)) = 4.3 h(1/0.6249 - 1/1.4826) = 4.3 h(1.600 - 0.6745) = 4.3 h(0.9255) = 4.3 h = 4.3 / 0.9255 ≈ 4.646 km

Now, substitute h back into the equation for PA: PA = h/tan(32°) PA = 4.646 / 0.6249 ≈ 7.436 km