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:

1. Initial State The eraser $$P$$ is at a fixed distance $$PC = \sqrt{5}$$ from the pivot $$C$$. The initial angle between the pencil and $$CP$$ is:$$\beta = \tan^{-1}(2)$$From the reference triangle:$$\tan \beta = 2$$ $$\sin \beta = \frac{2}{\sqrt{5}}$$ $$\cos \beta = \frac{1}{\sqrt{5}}$$

2. Final StateAfter rotating the pencil by an acute angle $$\alpha$$, the perpendicular distance from $$P$$ to the pencil is $$1$$ inch. Let the new angle between $$CP$$ and the pencil be $$\gamma$$.In the right-angled triangle formed:$$\sin \gamma = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{1}{\sqrt{5}}$$From this, we determine:$$\cos \gamma = \sqrt{1 - \sin^2 \gamma} = \sqrt{1 - \frac{1}{5}} = \frac{2}{\sqrt{5}}$$

$$\tan \gamma = \frac{\sin \gamma}{\cos \gamma} = \frac{1/\sqrt{5}}{2/\sqrt{5}} = \frac{1}{2}$$

3. Rotation CalculationFrom the diagram, the rotation angle $$\alpha$$ is the difference between the initial and final angles:$$\alpha = \beta - \gamma$$Apply the tangent subtraction identity:$$\tan \alpha = \tan(\beta - \gamma) = \frac{\tan \beta - \tan \gamma}{1 + \tan \beta \tan \gamma}$$Substitute the values:$$\tan \alpha = \frac{2 - \frac{1}{2}}{1 + (2 \cdot \frac{1}{2})} = \frac{1.5}{2} = \frac{3}{4}$$Final Answer:$$\alpha = \tan^{-1}\left(\frac{3}{4}\right)$$