Question

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Answer 1

Let the height the vertical tower be

OT = H m and OP = AB = x m

Given that, AP = 10 m

And ∠TPO = 60°, ∠TAB = 45°

Now, in ∆TPO,

tan 60° = `"OT"/"OP" = "H"/x`

⇒ `sqrt(3) = "H"/x`

⇒ `x = "H"/sqrt(3)`  ...(i)

And in ∆TAB,

tan 45° =  `"TB"/"AB" = ("H" - 10)/x`

⇒ 1 = `("H" - 10)/x`

⇒ `x = "H" - 10`

⇒ `"H"/sqrt(3) = "H" - 10`  ...[From equation (i)]

⇒ `"H" - "H"/sqrt(3)` = 10

⇒ `"H"(1 - 1/sqrt(3))` = 10

⇒ `"H"((sqrt(3) - 1)/sqrt(3))` = 10

⇒ H = `(10sqrt(3))/(sqrt(3) - 1)`

∴ H = `(10sqrt(3))/(sqrt(3) - 1) * (sqrt(3) + 1)/(sqrt(3) + 1)`  ...[By rationalisation]

= `(10sqrt(3)(sqrt(3) + 1))/(3 - 1)`

= `(10sqrt(3)(sqrt(3) + 1))/2`

⇒ H = `5sqrt(3)(sqrt(3) + 1) = 5(sqrt(3) + 3) "m"`.

Hence, the required height of the tower is `5(sqrt(3) + 3) "m"`.