A) 20 m and \[20\sqrt{3}\,m\]
B) 20 m and 60 m
C) 16 m and 48 m
D) None of these
Correct Answer: B
Solution :
\[\frac{H}{3}\cot \alpha =d\] and \[d=150\cot \varphi =60m\] or \[\frac{H}{3d}=\tan \alpha \] and \[\frac{H}{d}=\tan \beta \] \[\tan (\beta -\alpha )=\frac{1}{2}=\frac{\frac{H}{d}-\frac{H}{3d}}{1+\frac{{{H}^{2}}}{3{{d}^{2}}}}\] \[\Rightarrow \] \[1+\frac{{{H}^{2}}}{3{{d}^{2}}}=\frac{4H}{3d}\] \[\Rightarrow \] \[{{H}^{2}}-4dH+3{{d}^{2}}=0\] \[\Rightarrow \] \[{{H}^{2}}-80H+3\,(400)=0\] \[\Rightarrow \]\[H=20\]or \[60m\].You need to login to perform this action.
You will be redirected in
3 sec