Question

A vertical pole subtends an angle $ta{n}^{-1}\left(1/2\right)$ at a point P on the ground. The angle subtended by the upper half of the pole at the point P is

• A. $ta{n}^{-1}\left(1/4\right)$
• B. $ta{n}^{-1}\left(2/9\right)$
• C. $ta{n}^{-1}\left(1/8\right)$
• D. $ta{n}^{-1}\left(2/3\right)$

Answer 1

The correct option is B $ta{n}^{-1}\left(2/9\right)$

Let the pole AB subtend angle $\theta$ at P and the upper half BC of the pole subtend angle $\alpha$ at P below in the figure then tan $\theta =\frac{1}{2}=\frac{AB}{AP};tan(\theta -\alpha )=\frac{AC}{AP}=\frac{\left(1/2\right)AB}{AP}=\frac{1}{4}$
Now tan $\alpha =tan(\theta -(\theta -\alpha \left)\right)=\frac{tan\theta -tan(\theta -\alpha )}{1+tan\theta tan(\theta -\alpha )}=\frac{\frac{1}{2}-\frac{1}{4}}{1+\frac{1}{2}\times \frac{1}{4}}=\frac{2}{9}\Rightarrow \alpha =ta{n}^{-1}\frac{2}{9}$