JEE Main 2021 (Online) 3rd August Morning ShiftQuestion 1
A vertical pole consists of two portions, the lower being $\frac{1}{3}$rd of the whole. If the upper portion subtends an angle $\tan^{-1}\left(\frac{1}{2}\right)$ at a point on a horizontal plane drawn through the foot of the pole and at a distance $40$ m from it, then the height (in m) of the pole can be
Trigonometric Ratio & Identities
Solution
Concept. The angle subtended by the upper part is the difference of the elevations of the top and of the joint. Formulas used. (1) $\tan(u-v)=\dfrac{\tan u-\tan v}{1+\tan u\tan v}$Step 1 - Two elevations. With height $H$: $\tan u=\frac{H}{40}$ (top), $\tan v=\frac{H}{120}$ (joint at $\frac H3$). Step 2 - Use (1). $$\frac{\frac{H}{40}-\frac{H}{120}}{1+\frac{H^2}{4800}}=\frac{\frac{H}{60}}{1+\frac{H^2}{4800}}=\frac12.$$ Step 3 - Solve. $H^2-160H+4800=0$, so $H=40$ or $120$. Only $40$ is among the options. Answer: (C)