Question 405670: A motorist, traveling along a level highway at a speed of 60 km/h directly toward a mountain, observes that between 1:00 PM and 1:10 PM, the angle of elevation to the top of the mountain changes from 10 degrees to 70 degrees. Approximate the height of the mountain.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A motorist, traveling along a level highway at a speed of 60 km/h directly toward a mountain, observes that between 1:00 PM and 1:10 PM, the angle of elevation to the top of the mountain changes from 10 degrees to 70 degrees. Approximate the height of the mountain.
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First, let us find out the distance the motorist traveled between 1:00 and 1:10,travel time of 10 minutes.
distance=speed*time
=60km/hr*10minutes
10 minutes = 1/6 hr
distance=60*(1/6)=10km
Call this distance, d=10km
Call the height of the mountain,h
Call x, the distance the motorist would have traveled after 1:10 to reach the mountain.
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At 1:00 position
tan 10 deg=h/(x+d)
h=(x+d)tan 10 deg
=(x+10) tan 10 deg
At 1:10 position
tan 70 deg=h/x
h=(x)tan 70 deg
equating the two equations,
(x+10) tan 10 deg=(x)tan 70 deg
(x+10)/x=(tan 70 deg)/(tan 10 deg)=15.58
(x+10)/x=15.58
15.58x=x+10
14.58x=10
x=10/14.58=.686 km
h=x tan 70
=.686 tan70=1.88km
ans: The height of the mountain is 1.88 km
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