Question 635709: This is a trigonometry problem, but i am hoping you could also answer this...
The pilot in an airplane observes the angle of depression of a light directly below his line of flight to be 30.4 degrees. A minute later its angle of depression is 43 degrees. If he is flying horizontally in a straight course at the rate of 150 mi./hr., find the (a) altitude at which he is flying; (b) his distance from the light at the first point of observation.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The pilot in an airplane observes the angle of depression of a light directly below his line of flight to be 30.4 degrees. A minute later its angle of depression is 43 degrees. If he is flying horizontally in a straight course at the rate of 150 mi./hr., find the (a) altitude at which he is flying; (b) his distance from the light at the first point of observation.
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Draw the picture.
Fill in the angles you know from the problem description.
Draw an altitutde line from the 1st point of observation.
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Note: 150 mph = 150/60 = 5/2 = 2.5 miles per minute
Put that figure into the picture.
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You now have a triangle with angles 30.4, 137, and 12.6
and with side 2.5 miles
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Use the Law of Sines to find "his distance from the light
at the 1st point of observation".
s/sin(137)= 2.5/sin(12.6)
s = 7.82 miles
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Now you can find the altitude using:
alt = 7.82*sin(30.4) = 3.96 miles = 20,909 feet.
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Cheers,
Stan H.
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