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A tower $\mathrm{PQ}$ stands on a horizontal ground with base $\mathrm{Q}$ on the ground. The point Rdivides the tower in two parts such that $\mathrm{QR=15m}$ . If from a point $\mathrm{A}$ on the ground the angle of elevation of $\mathrm{R}$ is $\mathrm{60^{\circ}}$ and the part $\mathrm{PR}$ of the tower subtends an angle of $\mathrm{15^{\circ}}$ at $\mathrm{A}$ , then the height of the tower is :
Option: 1
$5(2 \sqrt{3}+3) \mathrm{m} \\$

Option: 2
$5(\sqrt{3}+3) \mathrm{m}$

Option: 3
$10(\sqrt{3}+1) \mathrm{m} \\$

Option: 4
$10(2 \sqrt{3}+1) \mathrm{m}$

Answers (1)

best_answer

$\mathrm{\frac{15}{A Q}=\tan 60^{\circ}}\\$ ........(1)

$\mathrm{\frac{15+x}{A Q}=\tan 75^{\circ}}\\$ ..........(2)

$\mathrm{\frac{(1)}{\text { (2) }} \Rightarrow x^{1}=10 \sqrt{3}}$

$\mathrm{so, P Q=5(2 \sqrt{3}+3) m}$

Hence correct option is 1

Posted by

Divya Prakash Singh

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