Watching from the top of a 30 m high tower,t | vedclass

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(C) Let the height of the tower be $AB = 30 \ m$. Let the stone be at point $C$ on the ground.The angle of depression from the top $A$ to the stone $C

Vedclass · Accepted Answer

$30$

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(C) Let the height of the tower be $AB = 30 \ m$. Let the stone be at point $C$ on the ground.The angle of depression from the top $A$ to the stone $C$ is $45^{\circ}$.Since the line of sight is parallel to the ground,the angle of elevation from the stone $C$ to the top $A$ is also $45^{\circ}$ (alternate interior angles).In the right-angled triangle $ABC$,we have:$	an(45^{\circ}) = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{AB}{BC}$.Since $	an(45^{\circ}) = 1$,we have $1 = \frac{30}{BC}$.Therefore,$BC = 30 \ m$.The distance of the stone from the base of the tower is $30 \ m$.

Watching from the top of a $30 \ m$ high tower,the angle of depression of a stone on the ground is found to be $45^{\circ}$. Then the distance of the stone from the base of the tower is $\ldots \ldots \ldots \ m$.

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