Trigonometrical Ratios And Identities Ques 5
- Two poles standing on a horizontal ground are of heights $5 \mathrm{m}$ and $10 \mathrm{m}$, respectively. The line joining their tops makes an angle of $15^{\circ}$ with the ground. Then, the distance (in $\mathrm{m}$ ) between the poles, is
(2019 Main, 9 April II)
(a) $5(\sqrt{3}+1)$
(b) $\frac{5}{2}(2+\sqrt{3})$
(c) $10(\sqrt{3}-1)$
(d) $5(2+\sqrt{3})$
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Answer:
Correct Answer: 5.(d)
Solution: (d) Given heights of two poles are $5 \mathrm{m}$ and $10 \mathrm{m}$.
i.e. from figure $A C=10 \mathrm{m}, D E=5 \mathrm{m}$
$\therefore \quad A B=A C-D E=10-5=5 \mathrm{m}$
Let $d$ be the distance between two poles.
Clearly, $\triangle A B E \sim \triangle A C F$
[by AA-similarity criterion]
$\therefore \quad \angle A E B=15^{\circ}$
In $\triangle A B E$, we have
$\tan 15^{\circ}=\frac{A B}{B E} \Rightarrow \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{5}{d}\quad \left [\because \tan 15^{\circ}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\right]$
$\Rightarrow \quad d=\frac{5(\sqrt{3}+1)}{(\sqrt{3}-1)}$
$ \begin{aligned} \Rightarrow \quad d & =5 \frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ & =\frac{5(3+2 \sqrt{3}+1)}{3-1}=\frac{5(2 \sqrt{3}+4)}{2} \\ & =\frac{2 \times 5(\sqrt{3}+2)}{2}=5(2+\sqrt{3}) \mathrm{m} \end{aligned} $
BETA
