The angle of elevation of a stationary hot air balloon from a point P on the level ground is 45^ . An observer moves a distance d meters from P towards the balloon along a straight path inclined at 30^ to the horizontal

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 100

Let the initial position of the observer be

P(0,0) and the balloon be at a height

h

.<br><br>Since the angle of elevation from

P

is

45^\circ

, the horizontal distance to the balloon is also

h

.<br><br>So, the coordinates of the balloon are

(h, h)

, where

h = 50(\sqrt{3}+1)

.<br><br>The observer moves a distance

d

along a path inclined at

30^\circ

to the horizontal.<br><br>The new position

Q

has coordinates \left(d \cos 30^\circ, d \sin 30^\circ\right) = \left(\dfrac{d\sqrt{3}}{2}, \dfrac{d}{2}\right)

.

From

Q

, the angle of elevation of the balloon is

60^\circ

.

The horizontal distance from

Q

to the balloon is

h - \dfrac{d\sqrt{3}}{2}

.

The vertical distance from

Q

to the balloon is

h - \dfrac{d}{2}

.

Thus,

\tan 60^\circ = \dfrac{h - \dfrac{d}{2}}{h - \dfrac{d\sqrt{3}}{2}} \Rightarrow \sqrt{3} = \dfrac{2h - d}{2h - d\sqrt{3}}

.

2h\sqrt{3} - 3d = 2h - d \Rightarrow 2d = 2h(\sqrt{3} - 1) \Rightarrow d = h(\sqrt{3} - 1)

.

Substitute

h = 50(\sqrt{3}+1)

:

d = 50(\sqrt{3}+1)(\sqrt{3}-1) = 50(3 - 1) = 100

m.

Answer:

100

The angle of elevation of a stationary hot air balloon from a point...

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