Exam

JEE Main

Year

2003

Subject

Mathematics

Chapter

Heights and Distances

JEE Main 2003 · Mathematics · Heights and Distances

The upper (3)/(4) th portion of a vertical pole subtends an angle ^...

The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $\tan ^{-1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance $40 \mathrm{~m}$ from the foot.

Step-by-step solution

$\tan \beta=\frac{\tan \theta-\tan \alpha}{1+\tan \theta \cdot \tan \alpha}$ or $\frac{3}{5}=\frac{\frac{\mathrm{h}}{40}-\frac{\mathrm{h}}{160}}{1+\frac{\mathrm{h}}{40} \cdot \frac{\mathrm{h}}{160}}$ $\mathrm{h}^2-200 \mathrm{~h}+6400=0, \mathrm{~h}=40$ or 160 metre Therefore possible height $=40$ metre