Two poles standing on a horizontal ground are of heights $5$ m and $10$ m respectively. The line joining their tops makes an angle of $15^{\circ}$ with ground. Then the distance (in m) between the poles, is :

\frac{5}{2} (2 + \sqrt{3})

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10 (\sqrt{3} - 1)

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5 (2 + \sqrt{3})

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5 (\sqrt{3} + 1)

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Solution

The correct option is C $5 (2 + \sqrt{3})$

Let $A B$ and $C D$ be the two towers. Distance between them is $x$ units.
$A E = 5$
$tan 15^{\circ} = \frac{A E}{E C} = \frac{5}{x}$
$\Rightarrow 2 - \sqrt{3} = \frac{5}{x}$
$\Rightarrow x = \frac{5}{2 - \sqrt{3}}$
or, $x = 5 (2 + \sqrt{3})$ units

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Two poles standing on a horizontal ground are of heights 5 m and 10 m respectively. The line joining their tops makes an angle of 15∘ with ground. Then the distance (in m) between the poles, is :

The correct option is C 5(2+√3) Let AB and CD be the two towers. Distance between them is x units. AE=5 tan15∘=AEEC=5x ⇒2−√3=5x ⇒x=52−√3 or, x=5(2+√3) units

Two poles standing on a horizontal ground are of heights $5$ m and $10$ m respectively. The line joining their tops makes an angle of $15^{\circ}$ with ground. Then the distance (in m) between the poles, is :

The correct option is C $5 (2 + \sqrt{3})$

Let $A B$ and $C D$ be the two towers. Distance between them is $x$ units.
$A E = 5$
$tan 15^{\circ} = \frac{A E}{E C} = \frac{5}{x}$
$\Rightarrow 2 - \sqrt{3} = \frac{5}{x}$
$\Rightarrow x = \frac{5}{2 - \sqrt{3}}$
or, $x = 5 (2 + \sqrt{3})$ units