SOLUTION: A hot-air balloon is rising vertically at a constant speed, an observer at a distant observes the elevation angle to be 30° at 10:00am, at 10:10am the elevation angle becomes 34°

Question 1189417: A hot-air balloon is rising vertically at a constant speed, an observer at a distant observes the elevation angle to be 30° at 10:00am, at 10:10am the elevation angle becomes 34°, then at 10:30am the elevation angle of the balloon should be closest (A) 34° (B) 39° (C) 41° (D) 42° (E) 43°
Answer by ikleyn(52797) (Show Source): You can put this solution on YOUR website!
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A hot-air balloon is rising vertically at a constant speed, an observer at a distant observes
the elevation angle to be 30° at 10:00am, at 10:10am the elevation angle becomes 34°,
then at 10:30am the elevation angle of the balloon should be closest
(A) 34° (B) 39° (C) 41° (D) 42° (E) 43°
~~~~~~~~~~~~~~~

Let d be the horizontal distance from the observer to the starting point of the balloon.
d is the constant in this problem.

Let h be the height of the balloon.  h = h(t) = vt is the linear function of time,
where v is the uniform (=constant) elevation rate in meters per minute.


At 10:00am (the reference time), we have h = tan(30°)*d =  = 0.57735d  (the referenced height).


At 10:10am, 10 minutes later, we have h = tan(34°)*d = 0.6745d.


Thus the elevation rate is  v =  = 0.009715d meters per minute.


At 10:30am, 30 minutes after 10:00am, the elevation will be  h = 0.57735d + 30v = 0.57735d + 30*0.009715d = 0.8688d.


At this time, the elevation angle  is   =  = 0.8688.


Hence,   = arctan(0.8688) = 40.98 degrees, or  41°  (rounded).


ANSWER.  Elevation angle will be close to 41°.

Solved.

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