An arc of `30^@` in one circle is double an arc in a second circle, the radius of which is three times the radius of the first. Then the angles subtended by the arc of the second circle at its centre is

Options

A. `3^@`
B. `4^@`
C. `5^@`
D. `6^@`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the relationship between the arcs and angles
We know that the angle subtended by an arc at the center of a circle is given by the formula:
\[ \text{Angle} = \frac{\text{Length of Arc}}{\text{Radius}} \]
From the problem, we have:
- An arc of \(30^\circ\) in the first circle.
- This arc is double the arc in the second circle.

### Step 2: Set up the equations for the first circle
Let the radius of the first circle be \(r\). The angle subtended by the arc in the first circle is \(30^\circ\).

Using the formula:
\[ 30^\circ = \frac{L_1}{r} \]
where \(L_1\) is the length of the arc in the first circle.

### Step 3: Relate the arcs of the two circles
Since the arc in the second circle is half of the arc in the first circle, we have:
\[ L_2 = \frac{L_1}{2} \]
where \(L_2\) is the length of the arc in the second circle.

### Step 4: Set up the equations for the second circle
The radius of the second circle is three times that of the first circle, so:
\[ \text{Radius of second circle} = 3r \]
Now, using the angle formula for the second circle, we have:
\[ \theta = \frac{L_2}{3r} \]

### Step 5: Substitute \(L_2\) into the equation
Substituting \(L_2 = \frac{L_1}{2}\) into the angle equation for the second circle:
\[ \theta = \frac{\frac{L_1}{2}}{3r} = \frac{L_1}{6r} \]

### Step 6: Relate \(L_1\) to the angle in the first circle
From the first circle, we know:
\[ L_1 = 30^\circ \cdot r \]
Substituting this into the equation for \(\theta\):
\[ \theta = \frac{30^\circ \cdot r}{6r} = \frac{30^\circ}{6} = 5^\circ \]

### Conclusion
Thus, the angle subtended by the arc of the second circle at its center is:
\[ \theta = 5^\circ \]