question_answer

The upper 3/4th portion of a vertical pole subtends an angle \[{{\tan }^{-1}}3/5\] at a point in the horizontal plane through its foot and at a distance 40 m from the foot. A possible height of the vertical pole is     AIEEE  Solved  Paper-2003

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 m