SOLUTION: The U.S. Weather Bureau defines a cloud ceiling as the altitude of the lowest clouds that cover more than half the sky. To determine a cloud ceiling, a powerful searchlight projec

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Question 220757: The U.S. Weather Bureau defines a cloud ceiling as the altitude of the lowest clouds that cover more than half the sky. To determine a cloud ceiling, a powerful searchlight projects a circle of light vertically on the bottom of the cloud. An observer sights the circle of light in the crosshairs of a tube called a clinometer. A pendant hanging vertically from the tube and resting on a protractor gives the angle of elevation. Find the cloud ceiling if the searchlight is located 1000 feet from the observer and the angle of elevation is 30.0� as measured with a clinometer at eye-height 6 feet. (Assume three significant digits.)
Please help solve and explain this.
Answer by ankor@dixie-net.com(22740) (Show Source):
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The U.S. Weather Bureau defines a cloud ceiling as the altitude of the lowest clouds that cover more than half the sky.
To determine a cloud ceiling, a powerful searchlight projects a circle of light vertically on the bottom of the cloud.
An observer sights the circle of light in the crosshairs of a tube called a clinometer.
A pendant hanging vertically from the tube and resting on a protractor gives the angle of elevation.
Find the cloud ceiling if the searchlight is located 1000 feet from the observer and the angle of elevation is 30.0� as measured with a clinometer at eye-height 6 feet.
(Assume three significant digits.)
Please help solve and explain this.
:
this is a right triangle, one leg is 1000 ft, the other leg is the elevation you are looking for.
Use the tangent of 30 degrees,
tan(30) =
h = 1000(.577345)
h = 577.350 ft
:
However since this triangle is at eye level (6'), elevation will be
577.35 + 6 = 583.35 ft