Question:
A chord of length 5 cm subtends an angle of 60° at the centre of a circle. The length, in cm, of a chord that subtends an angle of 120° at the centre of the same circle is
- 8
- 6√2
- 5√3
- \(2\pi \)
Correct Answer: 3
Since $\Delta \mathrm{OAB}$ is equilateral,
radius of the circle is 5 $\mathrm{cm}$ .
In $\Delta \mathrm{OCD},$ by sine rule,
$\frac{5}{\sin 30^{\circ}}=\frac{\mathrm{CD}}{\sin 120^{\circ}}$
$\Rightarrow \mathrm{CD}=5 \frac{\sqrt{3}}{2} \times 2$
$=5 \sqrt{3}$
Since $\Delta \mathrm{OAB}$ is equilateral,
radius of the circle is 5 $\mathrm{cm}$ .
In $\Delta \mathrm{OCD},$ by sine rule,
$\frac{5}{\sin 30^{\circ}}=\frac{\mathrm{CD}}{\sin 120^{\circ}}$
$\Rightarrow \mathrm{CD}=5 \frac{\sqrt{3}}{2} \times 2$
$=5 \sqrt{3}$
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