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 to (A) 34° (B) 39° (C) 41° (D)42° (E) 43°. Answer is C but I dont know why
let the distance between the observer and the point below the balloon be c
where c is a constant
let the height of the balloon at 10:00 be h
chart: let 10:00 am correspond with t = 0
let the angle be θ
tanθ = height/c
height = ctanθ
time height
0 c tan30 = .5773c
10 ctan34 = .6745c
30 ctan θ = x
since c is a constant, and the change in the height is also constant,
we can form a ratio:
(x - .5773)/(30-0) = (.6745 - .5773)/(10-0)
x - .5773 = 3(.0972)
x = .2916 + .5773
x = .8689
so tanθ = .8689
θ = arctan (.8689) = 40.98..°
= 41° to the nearest degree, answer C