Question

a pilot is flying over a straight highway. he determines the angles of depression to two mileposts, 6.5 mi apart, to be 36° and 51°, as shown in the figure. note: the picture is not drawn to scale. find the distance of the plane from point a. distance from a = mi find the elevation of the plane. height = mi enter your answer as a number; your answer should be accurate to 2 decimal places. question help: worked example 1

Explanation:

Step1: Find the third angle of the triangle

The sum of angles in a triangle is \(180^{\circ}\). The third angle \(\theta=180^{\circ}-36^{\circ}-51^{\circ}=93^{\circ}\)

Step2: Use the Law of Sines to find the distance from \(A\)

By the Law of Sines, \(\frac{d}{sin51^{\circ}}=\frac{6.5}{sin93^{\circ}}\), where \(d\) is the distance from \(A\). Then \(d = \frac{6.5\times sin51^{\circ}}{sin93^{\circ}}\)

$$d=\frac{6.5\times0.7771}{0.9986}\approx5.06$$

Step3: Find the elevation of the plane

Let \(h\) be the elevation. Using the right - triangle relationship (for the angle of depression \(36^{\circ}\) and the distance from \(A\) we just found), \(h = d\times sin36^{\circ}\)

$$h = 5.06\times0.5878\approx2.97$$

Answer:

distance from \(A = 5.06\) mi
height \(= 2.97\) mi