SOLUTION: 17 From the top of a vertical cliff 68 m high, an observer notices a yacht at sea. The angle of depression to the yacht is 47è. The yacht sails directly away from the cliff, and

Algebra ->  Trigonometry-basics -> SOLUTION: 17 From the top of a vertical cliff 68 m high, an observer notices a yacht at sea. The angle of depression to the yacht is 47è. The yacht sails directly away from the cliff, and      Log On


   



Question 714990: 17 From the top of a vertical cliff 68 m high, an observer
notices a yacht at sea. The angle of depression to the
yacht is 47è. The yacht sails directly away from the cliff,
and after 10 minutes the angle of depression is 15è. How
fast does the yacht sail?

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!

T = top of the cliff (we assume eye of observer is there, 68 meters above water surface)
B = bottom of the cliff (at water level)
X = initial position of the yacht
Y = position of the yacht 10 minutes later
TH = a horizontal line, parallel to BY.
angle HTX = 47%5Eo --> Angle BTX = 90%5Eo-47%5Eo=43%5Eo
angle HTY = 15%5Eo --> Angle BTY = 90%5Eo-15%5Eo=75%5Eo
BX=68m%2Atan%2843%5Eo%29=68m%2A0.93255=63.4m
BY=68m%2Atan%2875%5Eo%29=68m%2A3.732=253.8m
The yacht covered
253.8m-63.4m=190.4m in 10 minutes.
We could say that its speed in meters per minute is 190.4%2F10=19