posted by Emily .
From a ship off-shore, the angle of elevation of a hill is 1.1°. After the ship moves inland at 4.5 knots for 20 min, the angle of elevation is 1.4°. How high is the hill? (1 knot = 1 nautical mile = 6080 ft per hour)
As I was just about to get the answer, I realized that it was gonna be negative, which is definitely not correct. I apologize that I can't include a picture to clarify the problem a bit. Could someone maybe please explain how to get the answer? Any help is greatly appreciated! :)
Ok, let's make a diagram showing the sideview.
Let A be the origianl position of the ship, let B be the position after 20 minutes, Let P be the top of the hill, and Q be on AB extended so that BQ and PQ form a right angle.
angle Q = 90º, angle QBP = 1.4º, angle BAP = 1.1º
and AB = 4.5(20/60) = 1.5 nautical miles
By exterior angles, angle ABP = 178.6º
and angle APQ = .3 º
by sine law: BP/sin1.1 = 1.5/sin).3
5.49969 nautical miles or 33438 feet
now in right-angled triangle BQP
PQ/33438 = sin 1.4
PQ = 816.966 feet
the hill is appr. 817 feet high