UPSC NDA II 2015 Mathematics Paper

Concept:The distance between the poles can be found using the tangent of the angle formed by the line joining the tops with the horizontal.Explanation:Let the distance between the poles be $x$ metres.The vertical height difference between the tops is $20 - 10 = 10$ m.Consider the right triangle formed by the horizontal line and the line joining the tops.In this triangle, $\tan 15^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{10}{x}$.So, $x = \frac{10}{\tan 15^\circ}$.We know $\tan 15^\circ = 2 - \sqrt{3}$.Thus, $x = \frac{10}{2 - \sqrt{3}}$.Rationalise the denominator: multiply numerator and denominator by $2 + \sqrt{3}$:$x = \frac{10(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{10(2 + \sqrt{3})}{4 - 3} = 10(2 + \sqrt{3})$.Using $\sqrt{3} \approx 1.73$, we get $x = 10(2 + 1.73) = 10 \times 3.73 = 37.3$ m.Therefore, the distance is approximately $37.3$ m.Answer:The distance between the poles is approximately $37.3$ m, which corresponds to option B.