Question

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

• A. $2R \sin \frac{3\theta}{2} \sin \frac{\theta}{2}$
• B. $(1-\cos \frac{\theta}{2})(1+2\cos \frac{\theta}{2})R$
• C. $(1+\cos \frac{\theta}{2})(1-2\cos \frac{\theta}{2})R$
• D. $2R \sin \frac{3\theta}{4} \sin \frac{\theta}{4}$

Answer 1

Answer: (B), (D)

B, D

Answer 2

To find the perpendicular distance between the two parallel chords, we first determine their individual distances from the center of the circle.

Let the radius of the circle be $R$ and its center be $O$. Let the two parallel chords be $AB$ and $CD$. Since they are on the same side of the center, the longer chord will be closer to the center. The length of a chord is given by $2R \sin(\alpha/2)$, where $\alpha$ is the angle subtended by the chord at the center. The distance of a chord from the center is given by $R \cos(\alpha/2)$.

Let chord $AB$ subtend an angle $\theta$ at the center, so $\angle AOB = \theta$. Let chord $CD$ subtend an angle $2\theta$ at the center, so $\angle COD = 2\theta$.

The distance of chord $AB$ from the center, let's call it $d_{AB}$, is:

$d_{AB} = R \cos(\theta/2)$

The distance of chord $CD$ from the center, let's call it $d_{CD}$, is:

$d_{CD} = R \cos(2\theta/2) = R \cos(\theta)$

Since $\theta$ is an angle subtended by a chord, we assume $\theta \in (0, \pi)$. For $\theta \in (0, \pi)$, we have $\theta/2 \in (0, \pi/2)$. In this interval, $\theta/2 < \theta$, and the cosine function is decreasing. Therefore, $\cos(\theta/2) > \cos(\theta)$. This implies $d_{AB} > d_{CD}$. So, chord $AB$ is further from the center than chord $CD$.

The perpendicular distance between the two chords is the difference between their distances from the center, as they are on the same side:

Distance $= d_{AB} - d_{CD}$ Distance $= R \cos(\theta/2) - R \cos(\theta)$ Distance $= R (\cos(\theta/2) - \cos(\theta))$

Now, we need to compare this expression with the given options using trigonometric identities.

Option (B): $(1-\cos \frac{\theta}{2})(1+2\cos \frac{\theta}{2})R$

Let's expand this expression:

$R (1 \cdot 1 + 1 \cdot 2\cos(\theta/2) - \cos(\theta/2) \cdot 1 - \cos(\theta/2) \cdot 2\cos(\theta/2))$ $= R (1 + 2\cos(\theta/2) - \cos(\theta/2) - 2\cos^2(\theta/2))$ $= R (1 + \cos(\theta/2) - 2\cos^2(\theta/2))$

We know the double angle identity for cosine: $\cos(2x) = 2\cos^2(x) - 1$. So, $\cos(\theta) = 2\cos^2(\theta/2) - 1$. This implies $2\cos^2(\theta/2) = \cos(\theta) + 1$. Substitute this into the expression:

$= R (1 + \cos(\theta/2) - (\cos(\theta) + 1))$ $= R (1 + \cos(\theta/2) - \cos(\theta) - 1)$ $= R (\cos(\theta/2) - \cos(\theta))$

Thus, Option (B) is correct.

Option (D): $2R \sin \frac{3\theta}{4} \sin \frac{\theta}{4}$

We use the product-to-sum trigonometric identity: $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$. Let $A = \frac{3\theta}{4}$ and $B = \frac{\theta}{4}$.

Then $A-B = \frac{3\theta}{4} - \frac{\theta}{4} = \frac{2\theta}{4} = \frac{\theta}{2}$. And $A+B = \frac{3\theta}{4} + \frac{\theta}{4} = \frac{4\theta}{4} = \theta$.

So, $2R \sin \frac{3\theta}{4} \sin \frac{\theta}{4} = R (\cos(\theta/2) - \cos(\theta))$.

Thus, Option (D) is also correct.

Both options (B) and (D) are algebraically equivalent to the derived expression for the perpendicular distance between the chords.