Question

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Answer 1

$\huge{\mathcal{\underline{\underline{\color{indigo}{❥Answer}}}}}$

Question:-

From the top of a tower 100m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars.

Solution:-

Let the tower be PQ.

We have, PQ = 100m, ∠PQR = 30° and ∠PBQ = 45°

In ∆APQ,

$Tan \: 30° = \frac{PQ}{AP}$

$\implies{\frac{1}{\sqrt{3}} = \frac{100}{AP}}$

$\implies{AP = 100\sqrt{3} m}$

Also, in ∆BPQ,

$Tan \: 45° = \frac{PQ}{BP}$

$\implies{1 = \frac{100}{BP}}$

$\implies{BP = 100m}$

Now, AB = AP + BP

$\implies{100\sqrt{3} + 100}$

$\implies{100(\sqrt{3}+1)}$

$\implies{100(1.73 + 1)}$

$\implies{100 × 2.73}$

$\bold{\red{\implies{273m}}}$

Hence, the distance between the cars is 273m.

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