A) 40 metres
B) 20 metres
C) \[20\sqrt{3}metres\]
D) \[\frac{40}{3}\sqrt{3}metres\]
Correct Answer: B
Solution :
Let h be the height of pillar \[OB=h\cot {{30}^{o}}\] and \[OA=h\cot {{15}^{o}}\] Þ \[AB=OA-OB=h(\cot {{15}^{o}}-\cot {{30}^{o}})\] Þ \[\,h=\frac{40}{\cot {{15}^{o}}-\cot {{30}^{o}}}\] = 20 metre.You need to login to perform this action.
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