Question

A pilot in an airplane flying at 25,000 ft sees two towns directly ahead of her in a straight line. The angles of the depression to the towns are 25o and 50o , respectively. To the nearest foot, how far apart are the towns?

Answer 1

32635.17 ft

Step-by-step explanation:

In the diagram, AB = 25000 ft

Let C and D be the town, where CD = d (Distance between two towns)

By triangle, ABC

tan 50 = AB / AC

AC = 25000 / tan 50 = 20977.5 ft

By triangle, ABD

tan 25 = AB/AD

AD = 25000 / tan 25 = 53612.67 ft

So, the distance between two towns

d = AD - AC = 53612.67 - 20977.5 = 32635.17 ft

Thus, the distance between two towns is 32635.17 ft.

Answer 2

The towns are 32,635 ft apart.

Step-by-step explanation:

From the image drawn below:

AB = x ft

BC = a ft

AC = (x + a) ft

Considering triangle PCB,

tan 50° = 25000 / a

Or,

a = 25000 / tan 50°

Since tan x = 1/ cot x

a = 25000×cot 50°------------------------------------1

Considering triangle PCA,

tan 25° = 25000 / (a + x)

Or,

a + x = 25000 / tan 25°

Since tan x = 1/ cot x

a + x = 25000×cot 25° -------------------------------2

Thus, finding x from equation 1 and 2, we get:

x = 25000 (cot 25° - cot 50°)

Using cot 25° = 2.1445 and cot 50° = 0.8394, we get:

x ≈ 32,635 ft

Thus, the distance between two towns is 32,635 ft.