Question

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is `sqrt(st)`

Answer 1

Let the height of the tower is h.

And ∠ABC = θ

Given that, BC = s, PC = t

And angle of elevation on both positions are complementary.

i.e., ∠APC = 90° – θ  ...[If two angles are complementary to each other, then the sum of both angles is equal to 90°]

Now in ΔABC,

tan θ = `"AC"/"BC" = "h"/"s"`  ...(i)

And in ΔAPC,

tan(90° – θ) = `"AC"/"PC"`  ...[∵ tan(90° – θ) = cot θ]

⇒ cot θ = `"h"/"t"`

⇒ `1/tanθ = "h"/"t"`   `[∵ cot θ = 1/tan θ]` ...(ii)

On, multiplying equations (i) and (ii), we get

`tan θ * 1/tanθ = "h"/"s" * "h"/"t"`

⇒ `"h"^2/("st")` = 1

⇒ h² = st

⇒ h = `sqrt("st")`

So, the required height of the tower is `sqrt("st")`.

Hence proved.