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Two points $A$ and $B$ are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are $60 °$ and $45 °$ respectively. If the height of the tower is $15 m$ , then find the distance between these points.

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5 (3 - \sqrt{3}) m

5 (6 - \sqrt{3}) m

5 (4 - \sqrt{3}) m

5 (2 - \sqrt{3}) m

answer is A.

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Detailed Solution

Given that, angles of depression from top of tower are 60° and 45°.
Draw the diagram depicting the situation.
Question Image

In △ABC,

tanθ = \frac{Opposite}{Adjacent}

So,

\tan 60 ° = \frac{DC}{CA}

⇒

\sqrt{3}

\frac{15}{BC}

⇒

BC = \frac{15}{\sqrt 3}

⇒

BC = 5 \sqrt 3 m

Now, △ABD,

\tan 45 ° = \frac{AB}{BD}

\Rightarrow 1 = \frac{15}{BC + x}

[

\tan 45 ° = 1

]

\Rightarrow 1 = \frac{15}{5 \sqrt 3 + x}

\Rightarrow 5 \sqrt 3 + x = 15

\Rightarrow x = 15 - 5 \sqrt 3

\Rightarrow x = 5 (3 - \sqrt{3}) m

Thus, the distance between the two points C and D is

5 (3 - \sqrt{3}) m

.
Hence, option 1 is correct.

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