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 = 5 inches and ∠ PCB = ^ - 1 ( 2 ) . The acute angle through which the pencil must be rotated

Correct answer: A. ^ - 1 ((3)/(4)) From figure, sinβ = 1 5 tanβ = (1)/(2) ( α + β ) = 2 (α + β) = (tanα + (1)/(2))/(1 - tanα . (1)/(2)) = 2 α = ^ - 1 ((3)/(4))

JEE Main 202126 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main 2021 · 26 August · Evening Shift · Mathematics · Trigonometric Ratios & Identities

A 10 inches long pencil AB with mid point C and a small eraser P...

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 $∠ PCB = \tan^{- 1} ( 2 )$ . 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 :

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

Options

A $\tan^{- 1}\left(\dfrac{3}{4}\right)$
B $\tan^{- 1}\left(\dfrac{1}{2}\right)$
C $\tan^{- 1}\left(\dfrac{4}{3}\right)$
D $\tan^{- 1}\left(1\right)$

Correct answer

A. $\tan^{- 1}\left(\dfrac{3}{4}\right)$

Step-by-step solution

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 From figure,

$sin\beta = \dfrac{1}{\sqrt{5}}tan\beta = \dfrac{1}{2}$
$\tan ( \alpha + \beta ) = 2\tan \left(\alpha + \beta\right) = \dfrac{tan\alpha + \dfrac{1}{2}}{1 - tan\alpha . \dfrac{1}{2}} = 2$
$\alpha = \tan^{- 1}\left(\dfrac{3}{4}\right)$

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