Question

# A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30°$, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be $60°$. Find the time taken by the car to reach the foot of the tower from this point.

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Solution

## Step 1: Draw the diagram according to the question.Let $AB$ be the tower. $D$ is the initial and$C$ is the final position of the car respectively.As man is standing at the top of the tower so, Angles of depression are measured from $A$.$BC$ is the distance from the foot of the tower to the car.Step 2: Find relation between $BC$ and $CD.$From $\Delta ABC,\phantom{\rule{0ex}{0ex}}\mathrm{tan}60°=\frac{AB}{BC}$$\sqrt{3}=\frac{AB}{BC}\phantom{\rule{0ex}{0ex}}BC=\frac{AB}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}AB=\sqrt{3}BC$Again in $∆ABD$$\mathrm{tan}30°=\frac{AB}{BD}\phantom{\rule{0ex}{0ex}}\frac{1}{\sqrt{3}}=\frac{AB}{BD}\phantom{\rule{0ex}{0ex}}AB=\frac{BD}{\sqrt{3}}$Now, comparing Height $AB$, we get$\sqrt{3}BC=\frac{BD}{\sqrt{3}}$ (Since LHS are same, so RHS are also same)$3BC=BD\phantom{\rule{0ex}{0ex}}3BC=BC+CD\phantom{\rule{0ex}{0ex}}2BC=CDorBC=\frac{CD}{2}$Here, distance of $BC$ is half of $CD.$Step 3: Find the required time.Hence, the time taken is also half.Since, Time taken by car to travel distance $CD=6sec.$So, Time taken by car to travel $BC=\frac{6}{2}=3sec$.

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