Observing from the top of a 150, m high tower | vedclass

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Question

(A) Let the height of the tower be $AB = 150\, m$ and the position of the cycle be $C$.The angle of depression from the top $A$ to the cycle $C$ is $6

Vedclass · Accepted Answer

$50\sqrt{3}\, m$

Vedclass · Answer

(A) Let the height of the tower be $AB = 150\, m$ and the position of the cycle be $C$.The angle of depression from the top $A$ to the cycle $C$ is $60^{\circ}$,which implies the angle of elevation from $C$ to $A$ is also $60^{\circ}$ (alternate interior angles).In the right-angled triangle $ABC$,we have $	an(60^{\circ}) = \frac{	ext{Opposite}}{	ext{Adjacent}} = \frac{AB}{BC}$.Substituting the values: $\sqrt{3} = \frac{150}{BC}$.Therefore,$BC = \frac{150}{\sqrt{3}} = \frac{150\sqrt{3}}{3} = 50\sqrt{3}\, m$.Using $\sqrt{3} \approx 1.732$,we get $BC \approx 50 	imes 1.732 = 86.6\, m$.

Observing from the top of a $150\, m$ high tower,the angle of depression of a cycle on the ground is found to be $60^{\circ}$. How far away is the cycle from the tower?

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