A passenger in an airplane flying at an alti-
tude of 14 kilometers sees two towns directly
to the left of the plane. The angles of depres-
sion to the towns are 30◦ and 52◦.
How far apart are the towns?
Answer in units of ft.
Angle of depression is not enough, what azimuth bearings are they at? Maybe you mean at the same bearing...
draw the figure, let x be the distance apart, h the height of plane, a the distance from the close town to just below airplane.
h/a=tan52
h/(x+a)=tan30
You know h.
first equation, solve for a.
Put that in the second equation, solve for x.
To find the distance between the two towns, we can use trigonometry. Let's denote the angle of depression to the first town as θ1 (30°) and the angle of depression to the second town as θ2 (52°).
First, we need to determine the height from the plane to the ground. Since the plane is flying at an altitude of 14 kilometers, we convert it to the unit of feet. There are 3280.84 feet in one kilometer, so the altitude is 14 km * 3280.84 ft/km = 45,925.76 ft.
Let's assume the distance between the two towns is 'x.'
Now, let's consider the triangle formed by the plane, the first town, and the ground. We can apply the tangent function to this triangle:
tan(θ1) = opp/adj
tan(30°) = 45,925.76 ft / x
Similarly, for the triangle formed by the plane, the second town, and the ground, we have:
tan(θ2) = opp/adj
tan(52°) = 45,925.76 ft / (x + adj)
We can rearrange both equations to solve for 'x':
x = 45,925.76 ft / tan(30°)
x + adj = 45,925.76 ft / tan(52°)
Now, we solve for 'x':
x = 45,925.76 ft / tan(30°) ≈ 83,834.8 ft
x + adj = 45,925.76 ft / tan(52°) ≈ 71,888.7 ft
Therefore, the distance between the two towns is approximately 83,834.8 ft - 71,888.7 ft = 11,946.1 ft.