# Math Help - Height of Cloud Cover........

1. ## Height of Cloud Cover........

I don't remember how to do this problem. Can someone walk through it once so that I can refresh myself? Thanks!!

Height of Cloud Cover To measure the height of cloud cover at an airport, a worker shines a spotlight upward at an angle of 75 degrees from the horizontal. An observer 600 m away measures the angle of elevation to the spot of light to be 45 degrees. Find the height h of the cloud cover.

My book doesn't cover anything on how to do this and I don't remember when my instructor did it on the board. Thanks.........

2. Did you make a sketch?

From the first angle, you will get another angle of the triangle formed. It is 180 - 75 = 105 degrees. From there, you can find the last angle in the triangle. It gives 180 - (105+45) = 30 degrees.

(Or you use the property that the outside angle is equal to the sum of the two opposite angles in the triangle, and get 75 - 45 = 30 degrees)

Now, you can use the sine rule.

$\frac{L}{sin(45)} = \frac{600}{sin(30)}$

Solve for L, the distance from the first observer to the cloud, and you should get 848.5 m.

Now that you have that, draw a perpendicular line from the horizontal to the cloud. You get another triangle, this time a right angled triangle.

use the sine ratio to find the height of the cloud.

$sin(75) = \frac{h}{848.5}$

$h = 816.6 m = 817 m$ (3sf)