question_answer

A vertical pole (more than 100 m high) consists of two portions, the lower being one-third of the whole. If the upper portion subtends an angle \[{{\tan }^{-1}}\frac{1}{2}\]at a point in a horizontal plane through the foot of the pole and distance 40 ft from it, then the height of the pole is:

A)  100 ft                   

B)         120 ft   

C)         150 ft                   

D)         none of these

Correct Answer: B

Solution :

Let the  height of the pole BC be \[h.\] In \[\Delta \,ABC,\]                 \[\tan \beta =\frac{h}{40}\]                                ?(i) and in\[\Delta \Alpha \Beta D,\] \[\tan \alpha =\frac{h/3}{40}\] \[=\frac{h}{120}\]                                ?(ii)                 Now,       \[\tan \theta =\frac{1}{2}(given)\]                 \[\Rightarrow \]    \[\tan (\beta -\alpha )=\frac{1}{2}\] \[\Rightarrow \]    \[\frac{\tan \beta -\tan \alpha }{1+\tan \beta \tan \alpha }=\frac{1}{2}\] \[\Rightarrow \]    \[\frac{\frac{3h}{120}-\frac{h}{120}}{1+\frac{3{{h}^{2}}}{14400}}=\frac{1}{2}\]\[\Rightarrow \]\[h=120,40\] But \[h\]cannot be taken according to the given condition, therefore \[h=120\,ft.\]