Question

The angle of elevation of the top of a tower 30 m high, from two points on the level ground on its opposite sides are 45 degrees and 60 degrees. what is the distance between the two points?

Answer 1

distance between two points is 30+10√3m

Answer 2

The distance between the two points is 47.32m.

Let OT be to tower.

(The image shown to last of the question)

Therefore, Height of tower = OT = 30 m

Let A and B be the two points on the level ground on the opposite side of tower OT.

Then,

Angle of elevation from A = ∠ TAO = 45°

Angle of elevation from B = ∠ TBO = 60°

Distance between AB = AO + OB = x + y (say)

Now, in right triangle ATO,

tan 45° = OT/AO =30/x        

=> x =  30/tan45°  = 30 m,

In right triangle BTO

tan 60° =  OT/OB =30/y        

=> y = $\frac{30}{tan 60}$ =  $\frac{30}{\sqrt{3} }$ = $\frac{\sqrt[30]{3} }{3}$ =  17.32 m

Hence, the required distance = x + y = 30 + 17.32 = 47.32 m

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