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 :
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Setting up a coordinate system is often the easiest way to solve complex geometry problems. Place a key point at the origin and align an initial object with an axis. Then, use standard formulas for rotation and distance. The \(t = \tan(\theta/2)\) substitution is very useful for solving equations involving \(\sin\theta\) and \(\cos\theta\).
Step 1: Understanding the Geometry
Let's set up a coordinate system. Let the midpoint of the pencil C be at the origin (0,0). Initially, let the pencil lie along the x-axis, so B is at (5,0) and A is at (-5,0).
The eraser P is at a distance of \(\sqrt{5}\) from C. Let \(\theta = \angle PCB = \tan^{-1}(2)\). The coordinates of P are \((PC\cos\theta, PC\sin\theta)\).
Given \(\tan\theta = 2\), we can form a right triangle with opposite side 2 and adjacent side 1. The hypotenuse is \(\sqrt{2^2+1^2} = \sqrt{5}\).
So, \(\cos\theta = \frac{1}{\sqrt{5}}\) and \(\sin\theta = \frac{2}{\sqrt{5}}\).
The coordinates of P are \(\left(\sqrt{5} \cdot \frac{1}{\sqrt{5}}, \sqrt{5} \cdot \frac{2}{\sqrt{5}}\right) = (1, 2)\). Step 2: Rotation and Distance Formula
The pencil is rotated by an acute angle \(\alpha\) about C. The new line representing the pencil passes through the origin and has a slope of \(\tan\alpha\).
The equation of the rotated pencil is \(y = (\tan\alpha)x\), or \((\sin\alpha)x - (\cos\alpha)y = 0\).
The perpendicular distance from a point \((x_1, y_1)\) to a line \(Ax+By+C=0\) is \(d = \frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}\). Step 3: Detailed Calculation
We want the distance from P(1,2) to the line \((\sin\alpha)x - (\cos\alpha)y = 0\) to be 1.
\[ d = \frac{|(\sin\alpha)(1) - (\cos\alpha)(2)|}{\sqrt{\sin^2\alpha + (-\cos\alpha)^2}} = 1 \]
\[ |\sin\alpha - 2\cos\alpha| = 1 \]
This gives two possibilities:
1) \(\sin\alpha - 2\cos\alpha = 1\)
2) \(\sin\alpha - 2\cos\alpha = -1\)
We solve these using the substitution \(t = \tan(\alpha/2)\), where \(\sin\alpha = \frac{2t}{1+t^2}\) and \(\cos\alpha = \frac{1-t^2}{1+t^2}\). Case 1: \(\frac{2t}{1+t^2} - 2\frac{1-t^2}{1+t^2} = 1 \implies 2t - 2 + 2t^2 = 1+t^2 \implies t^2 + 2t - 3 = 0\).
\((t+3)(t-1) = 0 \implies t=1\) or \(t=-3\). Since \(\alpha\) is acute, \(\alpha/2\) is acute, so \(t=\tan(\alpha/2)>0\). Thus \(t=1 \implies \alpha/2 = 45^\circ \implies \alpha=90^\circ\). Case 2: \(\frac{2t}{1+t^2} - 2\frac{1-t^2}{1+t^2} = -1 \implies 2t - 2 + 2t^2 = -1-t^2 \implies 3t^2 + 2t - 1 = 0\).
\((3t-1)(t+1) = 0 \implies t=1/3\) or \(t=-1\). For acute \(\alpha\), we take \(t=1/3\).
If \(\tan(\alpha/2) = 1/3\), we find \(\tan\alpha\):
\[ \tan\alpha = \frac{2\tan(\alpha/2)}{1-\tan^2(\alpha/2)} = \frac{2(1/3)}{1-(1/3)^2} = \frac{2/3}{1-1/9} = \frac{2/3}{8/9} = \frac{2}{3} \cdot \frac{9}{8} = \frac{3}{4} \]
Step 4: Final Answer
The required acute angle is \(\alpha = \tan^{-1}\left(\frac{3}{4}\right)\).
