A) 50 metres
B) \[50\,\sqrt{3}\] metres
C) \[50\,\sqrt{2}\] metres
D) None of these
Correct Answer: D
Solution :
\[x=h\cot 3\alpha \] .....(i) \[(x+100)=h\cot 2\alpha \] ......(ii) \[(x+300)=h\cot \alpha \] ......(iii) From (i) and (ii), \[-100=h\,(\cot 3\alpha -\cot 2\alpha ),\] From (ii) and (iii), \[-200=h(\cot 2\alpha -\cot \alpha ),\] \[100=h\,\left( \frac{\sin \alpha }{\sin 3\alpha \sin 2\alpha } \right)\] and \[200=h\,\left( \frac{\sin \alpha }{\sin 2\alpha \sin \alpha } \right)\] or \[\frac{\sin 3\alpha }{\sin \alpha }=\frac{200}{100}\Rightarrow \frac{\sin 3\alpha }{\sin \alpha }=2\] \[\Rightarrow \] \[3\sin \alpha -4{{\sin }^{3}}\alpha -2\sin \alpha =0\] \[\Rightarrow \] \[4{{\sin }^{3}}\alpha -\sin \alpha =0\Rightarrow \sin \alpha =0\] or \[{{\sin }^{2}}\alpha =\frac{1}{4}={{\sin }^{2}}\left( \frac{\pi }{6} \right)\Rightarrow \alpha =\frac{\pi }{6}\] Hence, \[h=200\sin \frac{\pi }{3}=200\frac{\sqrt{3}}{2}=100\sqrt{3}\], {from (ii)} .You need to login to perform this action.
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