A) 20 m and \[20\sqrt{3}\]
B) 20 m and 60 m
C) 16 m and 48 m
D) None of these
Correct Answer: B
Solution :
[b] From the given figure we have \[\frac{H}{3d}=\tan \alpha \] and \[\frac{H}{d}=\tan \beta \] \[\tan (\beta -\alpha )=\frac{1}{2}=\frac{\frac{H}{d}-\frac{H}{3d}}{1+\frac{{{H}^{2}}}{3{{d}^{2}}}}\] \[\Rightarrow {{H}^{2}}-4dH+3{{d}^{2}}=0\] \[\Rightarrow {{H}^{2}}-80H+3(400)=0\] \[\Rightarrow H=20\] and 60 mYou need to login to perform this action.
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