Question

The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Answer 1

Let SQ = h be the tower.

∠SPQ = 30° and ∠SRQ = 60°

According to the question, 

The length of shadow is 50 m long the angle of elevation of the sun is 30° than when it was 60°.

So, PR = 50 m and RQ = x m

So in ∆SRQ, we have

tan 60° = `"h"/x`   ...`[∵ tan θ = "perpendicular"/"base" ⇒ tan 60^circ = "SQ"/"RQ"]`

⇒ `sqrt(3) = "h"/x`  ...`[∵ tan 60^circ = sqrt(3)]`

⇒ `x = "h"/sqrt(3)`

In ΔSPQ,

tan 30° = `"h"/(50 + x)`  ...`[∵ tan 30^circ = "SQ"/"PQ" = "SQ"/("PR" + "PQ")]`

⇒ `1/sqrt(3) = "h"/(50 + x)`  ...`[∵ tan 30^circ = 1/sqrt(3)]`

⇒ `50 + x = sqrt(3)"h"`

Substituting the value of x in the above equation, we get

⇒ `50 + "h"/sqrt(3) = sqrt(3)"h"`

⇒ `(50sqrt(3) + "h")/sqrt(3) = sqrt(3)"h"`

⇒ `50sqrt(3) + "h" = 3"h"`

⇒ `50sqrt(3) = 3"h" - "h"`

⇒ `3"h" - "h" = 50sqrt(3)`

⇒ 2h = `50sqrt(3)`

⇒ h = `(50sqrt(3))/2`

⇒ h = `25sqrt(3)`

Hence, the required height is `25sqrt(3)  "m"`.