A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of

Q: Why is $\alpha = \theta - \beta$?

Great question! Imagine the line $CP$ as a fixed vector. Initially, the pencil makes an angle $\theta$ with $CP$. After rotation, the pencil makes an angle $\beta$ with $CP$. The rotation angle $\alpha$ is simply the amount the pencil moved, which is the difference between the initial angle and the final angle.

Q: Can we use coordinate geometry instead of trigonometry?

Absolutely! You could place $C$ at the origin $(0,0)$, $P$ at $(\sqrt{5} \cos \theta, \sqrt{5} \sin \theta)$, and write the equation of the line. However, trigonometry is often faster and less prone to calculation errors in these types of problems. Always choose the path that minimizes algebraic complexity!

Q: Why did we assume $\beta$ is acute?

The problem asks for the acute angle of rotation. Furthermore, the perpendicular distance $d$ is a physical length, and in the right triangle formed, $\sin \beta = d/PC = 1/\sqrt{5}$. Since $1 < \sqrt{5}$, $\beta$ must be acute, which is consistent with our geometric setup.

Question

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 $\angle PCB = 	an^{-1}(2)$. 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:

Options:

(A) $	an^{-1}(3/4)$

(B) $	an^{-1}(1)$

(C) $	an^{-1}(4/3)$

(D) $	an^{-1}(1/2)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a pencil $AB$ resting on a table with a midpoint $C$. An eraser $P$ is positioned at a distance $PC = \sqrt{5}$ inches from the midpoint.

The initial angle between the line segment $CP$ and the pencil is defined by $	an 	heta = 2$. This establishes our coordinate reference for the static configuration.

### The Dynamic Shift

We rotate the pencil about its midpoint $C$ by an acute angle $\alpha$. The line $CP$ remains fixed, while the angle between $CP$ and the pencil changes to a new value, $\beta$.

The rotation angle $\alpha$ is defined as the difference between the initial angle and the new angle:
$$\alpha = 	heta - \beta$$

### The Trigonometric Bridge

The problem imposes a constraint: the perpendicular distance $d$ from the eraser $P$ to the rotated pencil must be exactly $1$ inch. By dropping a perpendicular from $P$ to the pencil, we form a right-angled triangle with hypotenuse $PC = \sqrt{5}$.

Using the definition of the sine function:
$$d = PC \sin \beta \Rightarrow 1 = \sqrt{5} \sin \beta$$

This yields $\sin \beta = \frac{1}{\sqrt{5}}$. Applying the Pythagorean identity, the adjacent side of this triangle is $\sqrt{(\sqrt{5})^2 - 1^2} = 2$.

Consequently, the tangent of the new angle is:
$$	an \beta = \frac{1}{2}$$

### The Grand Finale

We now possess the values $	an 	heta = 2$ and $	an \beta = \frac{1}{2}$. To find the rotation angle $\alpha$, we apply the tangent subtraction formula:
$$	an \alpha = 	an(	heta - \beta) = \frac{	an 	heta - 	an \beta}{1 + 	an 	heta 	an \beta}$$

Substituting the known values into the equation:
$$	an \alpha = \frac{2 - \frac{1}{2}}{1 + (2)(\frac{1}{2})}$$

Simplifying the numerator and the denominator:
$$	an \alpha = \frac{\frac{3}{2}}{1 + 1} = \frac{\frac{3}{2}}{2} = \frac{3}{4}$$

The final angle of rotation is **$\alpha = 	an^{-1}(\frac{3}{4})$**.

Select Answer:

Visualized Solution

Initial Configuration

Defining the Rotation

The Perpendicular Distance Condition

Substituting Known Values

Solving for $sin β$

Finding $tan β$

The Relation Between Angles

Applying the Compound Angle Formula

Substitution and Computation

Final Calculation

The Final Answer

The Sigma Insight: Properties of Inverse Trigonometric Functions

Analyzing the Setup

The Dynamic Shift

The Trigonometric Bridge

The Grand Finale

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