From the top of a building h metres high,the | vedclass

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(C) Let the height of the building be $h$ and the distance of the object from the foot of the building be $l$.According to the problem,the angle of de

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$h \cot 	heta$

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(C) Let the height of the building be $h$ and the distance of the object from the foot of the building be $l$.According to the problem,the angle of depression from the top of the building to the object is $	heta$.Since the line of sight is parallel to the ground,the angle of elevation from the object to the top of the building is also $	heta$ (alternate interior angles).In the right-angled triangle formed by the building and the ground,we have:$	an 	heta = \frac{	ext{Opposite side}}{	ext{Adjacent side}} = \frac{h}{l}$Rearranging the formula to solve for $l$:$l = \frac{h}{	an 	heta}$Since $\frac{1}{	an 	heta} = \cot 	heta$,we get:$l = h \cot 	heta$Therefore,the distance of the object from the foot of the building is $h \cot 	heta$.

From the top of a building $h$ metres high,the angle of depression of an object on the ground is $\theta$. The distance (in metres) of the object from the foot of the building is $\ldots \ldots \ldots$

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