Two parallel chords are drawn on the same side of the centre of a circle of radius R. It is found that they subtend and angle of `theta` and `2theta` at the centre of the circle. The perpendicular distance between the chords is

Options

A. `2Rsin. (3theta)/(2) sin. (theta)/(2)`
B. `(1-cos. (theta)/(2))(1+2 cos. (theta)/(2))R`
C. `(1+cos. (theta)/(2))(1-2cos. (theta)/(2))R`
D. `2 R sin. (3theta)/(4) sin. (theta)/(4)`

Answer

To find the perpendicular distance between two parallel chords in a circle that subtend angles of \( \theta \) and \( 2\theta \) at the center, we can follow these steps:

### Step 1: Understand the Geometry
We have a circle with center \( O \) and radius \( R \). Two parallel chords \( AB \) and \( CD \) are drawn such that chord \( AB \) subtends an angle of \( 2\theta \) at the center and chord \( CD \) subtends an angle of \( \theta \).

### Step 2: Identify the Right Triangles
For chord \( AB \):
- The angle subtended at the center is \( 2\theta \).
- The perpendicular distance from the center \( O \) to chord \( AB \) is denoted as \( OP \).
- In triangle \( OAP \) (where \( A \) is one endpoint of chord \( AB \)), we have:
\[ \cos(\theta) = \frac{OP}{R} \implies OP = R \cos(\theta) \]

For chord \( CD \):
- The angle subtended at the center is \( \theta \).
- The perpendicular distance from the center \( O \) to chord \( CD \) is denoted as \( OQ \).
- In triangle \( OCQ \) (where \( C \) is one endpoint of chord \( CD \)), we have:
\[ \cos\left(\frac{\theta}{2}\right) = \frac{OQ}{R} \implies OQ = R \cos\left(\frac{\theta}{2}\right) \]

### Step 3: Calculate the Perpendicular Distance Between the Chords
The perpendicular distance \( PQ \) between the two chords \( AB \) and \( CD \) is given by:
\[ PQ = OQ - OP \]
Substituting the expressions for \( OQ \) and \( OP \):
\[ PQ = R \cos\left(\frac{\theta}{2}\right) - R \cos(\theta) \]
Factoring out \( R \):
\[ PQ = R \left( \cos\left(\frac{\theta}{2}\right) - \cos(\theta) \right) \]

### Step 4: Use the Cosine Difference Formula
Using the formula for the difference of cosines:
\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \]
Let \( A = \frac{\theta}{2} \) and \( B = \theta \):
\[ PQ = R \left( -2 \sin\left(\frac{\frac{\theta}{2} + \theta}{2}\right) \sin\left(\frac{\frac{\theta}{2} - \theta}{2}\right) \right) \]
This simplifies to:
\[ PQ = R \left( -2 \sin\left(\frac{3\theta}{4}\right) \sin\left(-\frac{\theta}{4}\right) \right) \]
Since \( \sin(-x) = -\sin(x) \):
\[ PQ = 2R \sin\left(\frac{3\theta}{4}\right) \sin\left(\frac{\theta}{4}\right) \]

### Final Result
Thus, the perpendicular distance between the two parallel chords is:
\[ PQ = 2R \sin\left(\frac{3\theta}{4}\right) \sin\left(\frac{\theta}{4}\right) \]