Prove that height of the tower -tan-tan-tan-tan-

Let h m be the height of tower Given DC = am and BC = xm In Δ ABC tan β = h/x ⇒ x = h/tan β ...(1) In Δ ABD, we have tan α = h/a + x ⇒ tan α = h/a + h/tan β [u ...

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Standard XII

Physics

Relative Motion: Rain Example

Prove that he...

Question

Prove that height of the tower = $\frac{α t a n β t a n α}{t a n β - t a n α}$

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Solution

Let h m be the height of tower
Given DC = am
and BC = xm

In $Δ A B C$
$t a n β = \frac{h}{x}$
$\Rightarrow x = \frac{h}{t a n β} . . . (1)$
In $Δ A B D,$ we have
$t a n α = \frac{h}{a + x}$
$\Rightarrow t a n α = \frac{h}{a + \frac{h}{t a n β}} [u s i n g (1)]$
$\Rightarrow t a n α = \frac{h t a n β}{α t a n β + h}$
$\Rightarrow α t a n β t a n α + h t a n α = h t a n β$
$\Rightarrow h (t a n β - t a n α) = a t a n β t a n α$
$\Rightarrow h = \frac{α t a n β t a n α}{t a n β - t a n α}$

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Prove that height of the tower = αtanβtanαtanβ−tanα

Let h m be the height of tower Given DC = am and BC = xm In ΔABC tanβ=hx ⇒x=htanβ...(1) In ΔABD, we have tanα=ha+x ⇒tanα=ha+htanβ[using(1)] ⇒tanα=h tanβαtan β+h ⇒αtanβ tanα+h tan α=h tan β ⇒h(tanβ−tanα)=a tan β tan α ⇒h=αtanβtanαtanβ−tanα

Prove that height of the tower = $\frac{α t a n β t a n α}{t a n β - t a n α}$