JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
A $10$ inches long pencil $AB$ with mid point $C$ and a small eraser $P$ are placed on the horizontal top of a table such that $PC=\sqrt{5}$ inches and $\angle PCB=\tan^{-1}(2)$. The acute angle through which the pencil must be rotated about $C$ so that the perpendicular distance between eraser and pencil becomes exactly $1$ inch is : [FIGURE NOT TRANSCRIBED: pencil $AB$ with $AC=CB=5$ in and point $P$ at distance $\sqrt{5}$ in from $C$; see the paper image]
Trigonometric Ratio & Identities
Solution
Concept. Put $C$ at the origin with $CB$ on the $x$-axis; then $P=(1,2)$. After turning the pencil through $\varphi$, the distance from $P$ to the line $CB'$ must be $1$.Formulas used. (1) Distance from $(x_0,y_0)$ to the line through $O$ with direction $(\cos\varphi,\sin\varphi)$: $|x_0\sin\varphi-y_0\cos\varphi|$Step 1 - Position of $P$. $PC=\sqrt5$ and $\tan\angle PCB=2$, so $P=(1,2)$. Step 2 - Condition. By (1), $|\sin\varphi-2\cos\varphi|=1$. Step 3 - Solve. Take $\sin\varphi=2\cos\varphi-1$. Squaring with $\sin^2\varphi+\cos^2\varphi=1$: $5\cos^2\varphi-4\cos\varphi=0$, so $\cos\varphi=\frac45$, $\sin\varphi=\frac35$ (the other case $\cos\varphi=0$ gives $90^\circ$). The other sign gives $\cos\varphi\le0$, not acute. Step 4 - Angle. $\varphi=\tan^{-1}\frac34$. Answer: (B)