Watching from the top of the x m high buildi | vedclass

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(D) Let $AB$ be the building of height $x \ m$ and $C$ be the position of the child on the ground.The angle of depression from the top $A$ to the chil

Vedclass · Accepted Answer

$x \cot 	heta$

Vedclass · Answer

(D) Let $AB$ be the building of height $x \ m$ and $C$ be the position of the child on the ground.The angle of depression from the top $A$ to the child at $C$ is $	heta$.Since the horizontal line at $A$ is parallel to the ground $BC$,the angle of elevation from $C$ to $A$ is also $	heta$ (alternate interior angles).In the right-angled triangle $	riangle ABC$,we have:$	an 	heta = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{AB}{BC}$$	an 	heta = \frac{x}{BC}$Therefore,$BC = \frac{x}{	an 	heta} = x \cot 	heta \ m$.Thus,the distance of the child from the base of the building is $x \cot 	heta \ m$.

Watching from the top of the $x \ m$ high building,the angle of depression of a child on the ground is found to be $\theta$. Then,the distance of the child from the base of the building is..........

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