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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)`
Solution
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.
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