Let XY be the diameter of a semicircle with centre O. Let A be a variable point on the semicircle and B another point on the semicircle such that AB is parallel to XY. The value of `angleBOY` for which the inradius of triangle AOB is maximum, is

Options

A. `cos^(-1)((sqrt(5)-1)/(2))`
B. ` sin^(-1)((sqrt(5)-1)/(2))`
C. `(pi)/(3)`
D. `(pi)/5`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have a semicircle with diameter XY and center O. A point A is on the semicircle, and point B is another point on the semicircle such that line segment AB is parallel to XY.

### Step 2: Define the Angles
Let the angle \( \angle BOY = \theta \). We will express the inradius of triangle AOB in terms of \( \theta \).

### Step 3: Identify the Triangle Area
The area \( A \) of triangle AOB can be expressed as:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( AB \) can be expressed in terms of \( r \) (the radius of the semicircle) and \( \theta \).

### Step 4: Express Base and Height
1. The height from O to line AB is \( r \sin \theta \).
2. The base \( AB \) can be calculated as \( 2r \cos \theta \) since \( AB \) is parallel to the diameter and spans the width across the semicircle.

### Step 5: Calculate the Area
Substituting these values into the area formula:
\[ \text{Area} = \frac{1}{2} \times (2r \cos \theta) \times (r \sin \theta) = r^2 \sin \theta \cos \theta \]

### Step 6: Calculate the Semi-perimeter
The semi-perimeter \( s \) of triangle AOB is given by:
\[ s = \frac{OA + OB + AB}{2} = \frac{r + r + 2r \cos \theta}{2} = r(1 + \cos \theta) \]

### Step 7: Find the Inradius
The inradius \( r_i \) can be expressed as:
\[ r_i = \frac{\text{Area}}{s} = \frac{r^2 \sin \theta \cos \theta}{r(1 + \cos \theta)} = \frac{r \sin \theta \cos \theta}{1 + \cos \theta} \]

### Step 8: Maximize the Inradius
To find the maximum inradius, we need to differentiate \( r_i \) with respect to \( \theta \) and set the derivative equal to zero:
\[ \frac{d}{d\theta} \left( \frac{\sin \theta \cos \theta}{1 + \cos \theta} \right) = 0 \]

### Step 9: Apply the Quotient Rule
Using the quotient rule:
\[ \frac{d}{d\theta} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \]
where \( u = \sin \theta \cos \theta \) and \( v = 1 + \cos \theta \).

### Step 10: Solve the Derivative Equation
After differentiating and simplifying, we will find the critical points for \( \theta \).

### Step 11: Find the Value of \( \theta \)
Solving the resulting equation will yield the value of \( \theta \) for which the inradius is maximized.

### Final Step: Conclusion
The value of \( \angle BOY \) for which the inradius of triangle AOB is maximum is:
\[ \theta = \cos^{-1} \left( \frac{\sqrt{5} - 1}{2} \right) \]