(D) Let the height of the object be $h$. Let $B$ be the point on the ground directly below the object $A$, and $C$ and $D$ be the two points on the gr

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

(D) Let the height of the object be $h$. Let $B$ be the point on the ground directly below the object $A$, and $C$ and $D$ be the two points on the ground such that $CD = d$ and $BD = x$.In $\triangle ABC$, $\tan \alpha = \frac{h}{x+d} \Rightarrow x+d = h \cot \alpha$ $(i)$In $\triangle ABD$, $\tan \beta = \frac{h}{x} \Rightarrow x = h \cot \beta$ (ii)Substituting the value of $x$ from (ii) into $(i)$:$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 Mathematics Trigonometrical Equations Height and Distance

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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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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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