From the top of a hill h meters high- the angle of depression of the top and the bottom of a pillar are - - respectively- Then the heightin meters of the pillar is

The correct option is A h(tanβ tanα)tanβ tanα #160; = h da tanβ = ha tanαtanβ = h dh ⇒tanαtanβ = #160; 1 dh ⇒dh= tanβ tanαt ...

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

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Question

From the top of a hill $h$ meters high, the angle of depression of the top and the bottom of a pillar are $α, β$ respectively. Then the height(in meters) of the pillar is

\frac{h (tan β - tan α)}{tan β}

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\frac{h (tan α - tan β)}{tan α}

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\frac{h (tan β + tan α)}{tan β}

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\frac{h (tan β + tan α)}{tan α}

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Solution

The correct option is A $\frac{h (tan β - tan α)}{tan β}$
$tan α = \frac{h - d}{a}$
$tan β = \frac{h}{a}$
$\frac{tan α}{tan β} = \frac{h - d}{h}$
$\Rightarrow \frac{tan α}{tan β} = 1 - \frac{d}{h}$
$\Rightarrow \frac{d}{h} = \frac{tan β - tan α}{tan β}$
$\Rightarrow d = h (\frac{tan β - tan α}{tan β})$

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From the top of a hill h meters high, the angle of depression of the top and the bottom of a pillar are α, β respectively. Then the height(in meters) of the pillar is

The correct option is A h(tanβ−tanα)tanβtanα=h−datanβ=ha tanαtanβ=h−dh⇒tanαtanβ=1−dh⇒dh=tanβ−tanαtanβ⇒d=h(tanβ−tanαtanβ)

From the top of a hill $h$ meters high, the angle of depression of the top and the bottom of a pillar are $α, β$ respectively. Then the height(in meters) of the pillar is