In Fig. 3, $AB$ is tower of height $50$ m. A man standing on its top, observes two cars on the opposite sides of the tower with angles of depression $30°$ and $45°$ respectively. Find the distance bet

Question

In Fig. 3, $AB$ is tower of height $50$ m. A man standing on its top, observes two cars on the opposite sides of the tower with angles of depression $30°$ and $45°$ respectively. Find the distance between the two cars.

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Accepted Answer

Step 1: $	an 30° = \frac{1}{\sqrt{3}} = \frac{AB}{BC} \Rightarrow BC = AB\sqrt{3} = 50\sqrt{3}$ m ... (i)
Step 2: $	an 45° = \frac{AB}{BD} = 1 \Rightarrow BD = AB = 50$ m ... (ii)
Step 3: $CD = BC + BD = 50\sqrt{3} + 50 = 50(\sqrt{3} + 1)$ m

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