(D) Let the height of the object be $h$.In $\triangle ABC$,we have $\tan \alpha = \frac{h}{x+d}$,which implies $x+d = h \cot \alpha$ $(i)$.In $\triang

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$\frac{d}{\cot \alpha-\cot \beta}$

(D) Let the height of the object be $h$.In $\triangle ABC$,we have $\tan \alpha = \frac{h}{x+d}$,which implies $x+d = h \cot \alpha$ $(i)$.In $\triangle ABD$,we have $\tan \beta = \frac{h}{x}$,which implies $x = h \cot \beta$ (ii).Substituting the value of $x$ from (ii) into $(i)$,we get:$h \cot \beta + d = h \cot \alpha$$d = h \cot \alpha - h \cot \beta$$d = h(\cot \alpha - \cot \beta)$$h = \frac{d}{\cot \alpha - \cot \beta}$

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object,the angle of elevation is found to be $\beta$. Then the height (in metres) of the object is

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A
$d \tan \alpha$
B
$d \cot \beta$
C
$\frac{d}{\cot \alpha+\cot \beta}$
D
$\frac{d}{\cot \alpha-\cot \beta}$

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The angle of elevation of an object from a point $P$ on the level ground is $\alpha$. Moving $d$ metres on the ground towards the object,the angle of elevation is found to be $\beta$. Then the height (in metres) of the object is

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