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Question

(A) Let the height of the tower be $AB = 50\, m$. Let the two cars be at points $C$ and $D$ such that the angle of depression from the top $A$ to $C$

Vedclass · Accepted Answer

$57.73$

Vedclass · Answer

(A) Let the height of the tower be $AB = 50\, m$. Let the two cars be at points $C$ and $D$ such that the angle of depression from the top $A$ to $C$ is $30^{\circ}$ and to $D$ is $60^{\circ}$.In $	riangle ABD$,$	an(60^{\circ}) = \frac{AB}{BD} \implies \sqrt{3} = \frac{50}{BD} \implies BD = \frac{50}{\sqrt{3}} \approx 28.87\, m$.In $	riangle ABC$,$	an(30^{\circ}) = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{50}{BC} \implies BC = 50\sqrt{3} \approx 86.60\, m$.The distance between the two cars is $CD = BC - BD = 50\sqrt{3} - \frac{50}{\sqrt{3}} = 50 \left( \frac{3-1}{\sqrt{3}} ight) = \frac{100}{\sqrt{3}} \approx 57.73\, m$.

The angles of depression of two cars parked in the same direction of a $50\, m$ high tower are found to be $30^{\circ}$ and $60^{\circ}$. Find the distance between these two cars. (in $, m$)

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