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प्रश्न
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.
उत्तर
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"`.
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