A) 20m
B) 40m
C) 60m
D) 80 m
Correct Answer: B
Solution :
Given, \[{{\theta }_{2}}={{\tan }^{-1}}\frac{3}{5}\] \[\tan {{\theta }_{2}}=\frac{3}{5}\] ... (i) In \[\Delta AOC\], \[\tan {{\theta }_{1}}=\frac{AC}{OA}\] \[=\frac{\frac{1}{4}h}{40}=\frac{h}{60}\] ?. (ii) In \[\Delta AOB\], \[\tan \,({{\theta }_{1}}+{{\theta }_{2}})=\frac{AB}{OA}=\frac{h}{40}\] \[\Rightarrow \] \[\frac{\tan {{\theta }_{1}}+\tan {{\theta }_{2}}}{1-\tan {{\theta }_{1}}\tan {{\theta }_{2}}}=\frac{h}{40}\] \[\Rightarrow \] \[\frac{\frac{h}{40}+\frac{3}{5}}{1-\frac{h}{160}\times \frac{3}{5}}=\frac{h}{40}\] [from Eqs.(i) and (ii)] \[\Rightarrow \] \[\frac{5\,(h+96)}{800-3h}=\frac{h}{40}\] \[\Rightarrow \] \[200\,(h-96)=800h-3{{h}^{2}}\] \[\Rightarrow \] \[200h+19200=800h-3{{h}^{2}}\] \[\Rightarrow \] \[3{{h}^{2}}-600h+19200=0\] \[\Rightarrow \] \[{{h}^{2}}-200h+6400=0\] \[\Rightarrow \] \[(h-160)\,(h-40)=0\] \[\Rightarrow \] h = 160 or h = 40 \[\therefore \] Height of the vertical pole = 40 mYou need to login to perform this action.
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