As observed from the top of a building,the an | vedclass

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(B) Let $\overline{AB}$ be the building and $C$ be the position of the man on the ground.Given that the distance $BC = 10 \, m$.The angle of depressio

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$10$

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(B) Let $\overline{AB}$ be the building and $C$ be the position of the man on the ground.Given that the distance $BC = 10 \, m$.The angle of depression from the top $A$ to the man at $C$ is $45^{\circ}$.Since the line of sight is parallel to the ground,the angle of elevation from $C$ to $A$ is equal to the angle of depression,so $\angle ACB = 45^{\circ}$.In the right-angled triangle $\Delta ABC$,we have:$	an(C) = \frac{	ext{Opposite side}}{	ext{Adjacent side}} = \frac{AB}{BC}$$	an(45^{\circ}) = \frac{AB}{10}$Since $	an(45^{\circ}) = 1$,we have:$1 = \frac{AB}{10}$$AB = 10 \, m$.Thus,the height of the building is $10 \, m$.

As observed from the top of a building,the angle of depression of a man standing $10 \, m$ away from the building is found to be $45^{\circ}$. Find the height of the building (in $m$).

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