Question

The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

Answer 1

Let the height of the tower be h.

Also, SR = x m,

Given that, QS = 20 m

`sqrt("PQR")` = 30°

And `sqrt("PSR") = sqrt("PQR") + 15^circ`

= 30° + 15°

= 45°

Now, In ∆PSR,

tan 45° = `"PR"/"SR" = "h"/x`

⇒ `1 = "h"/x`  ...[∵ tan 45° = 1]

⇒ x = h  ...(i)

Now, In ∆PQR

tan 30° = `"PR"/"QR" = "PR"/("QS" + "SR")` 

⇒ tan 30° = `"h"/(20 + x)`

⇒ 20 + x = `"h"/(tan 30^circ) = "h"/(1/sqrt(3))`

⇒ 20 + x = `"h"sqrt(3)`

⇒ 20 + h = `"h"sqrt(3)`  ...[From (i)]

⇒ 20 = `"h"sqrt(3) - "h"`

⇒ `"h"(sqrt(3) - 1)` = 20

⇒ h = `20/(sqrt(3) - 1) * (sqrt(3) + 1)/(sqrt(3) + 1)`  ...[By rationalisation]

⇒ h = `(20(sqrt(3) + 1))/(3 - 1) = (20(sqrt(3) + 1))/2`

⇒ h = `10(sqrt(3) + 1) "m"` 

Hence, the required height of tower is `10(sqrt(3) + 1) "m"`.