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EEJ MAIN Mathematics QUESTION #7461
Question 1
Two poles standing on horizontal ground are of heights 5 m and 10 m respectively. The line joining their tops makes an angle of 15° with the ground. Then the distance (in m) between the poles is:
  • \(5(2+\sqrt{3})\)✔️
  • \(5(\sqrt{3}+1)\)
  • \(\frac{5}{2}(2+\sqrt{3})\)
  • \(10(\sqrt{3}-1)\)
Correct Answer Explanation
Let the distance between poles be \(d\). The height difference is \(10-5=5\) m. Using trigonometry: \(\tan(15°) = \frac{5}{d}\). We know \(\tan(15°) = 2-\sqrt{3}\) (from the formula \(\tan(45°-30°) = \frac{\tan 45° - \tan 30°}{1+\tan 45° \tan 30°} = \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}\)). Therefore, \(d = \frac{5}{2-\sqrt{3}} = \frac{5(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})} = \frac{5(2+\sqrt{3})}{4-3} = 5(2+\sqrt{3})\).