Is the shadow of the tower is 30 m long when the suns elevation is 30 degree What is the length of the shadow when the suns elevation is 60 degree?

Let SQ = h be the tower.

∠SPQ = 30° and ∠SRQ = 60°

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

So, PR = 50 m and RQ = x m

`tan 60^circ = h/x`   .......`[because tan theta = "perpendicular"/"base" ⇒ tan 60^circ = (SQ)/(RQ)]`

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

⇒ ` x = h/sqrt(3)`

In ΔSPQ,

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

⇒ `1/sqrt(3) = h/(50 + x)`  .....`[because 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 = 3h`

⇒ `50sqrt(3) = 3h - h`

⇒ `3h - h = 50sqrt(3)`

⇒ `2h = 50sqrt(3)`

⇒ `h = (50sqrt(3))/2`

⇒ `h = 25sqrt(3)`

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