Question

A 1.5 m tall boy is standing at some distance from a 31.5 m tall building. If he walks ’d’ m towards the building the angle of elevation of the top of the building changes from 30° to 60°. Find the length d. (Take `sqrt3 = 1.73`)

Options

• 30.15 m

• 38.33m

• 22.91m

• 34.55m

Answer 1

34.55m

Explanation:

The above figure represents the situation given in question

AA' = BB' = CC' = 1.5

`"tan" (angle "DB'C'") = "DC'"/"B'C'"`

and DC' = DC - CC' = 31.5 - 1.5 = 30 m

`therefore "tan"  60^circ = 30/"B'C'"`

`=> "B'C'" = 30/("tan"  60^circ) = 30/sqrt3`  ...(1)

Similarly,

`"tan"  30^circ = "DC'"/"A'C'" = "DC'"/("A'B'" + "B' C'")`

`=> "A'B'" + "B'C'" = "DC'"/("tan"  30^circ) = 30 sqrt3` ....(2)

Subtracting eq (1) from (2)

`"A'B'" = 30 sqrt3 - 30/sqrt3 = 30 (sqrt3 - 1/sqrt3) = 34.55  "m"`