A chord of length $5 \mathrm{~cm}$ subtends an angle of $60^{\circ}$ at the centre of a circle. The length, in cm , of a chord that subtends an angle of $120^{\circ}$ at the centre of the same circle is

When a chord subtends a 60° angle at the center, something special happens!

The chord and the two radii form a triangle. Since both radii are equal (let's call the radius r), we have an isosceles triangle with:

Two sides of length r (the radii)

One side of length 5 cm (the chord)

The angle between the radii is 60°

Here's the key insight: In an isosceles triangle where the angle between the equal sides is 60°, all three sides must be equal! This makes it an equilateral triangle.

The other two angles must each be (180° - 60°)/2 = 60°. So all angles are 60°, making it equilateral.

Therefore: radius = chord length = 5 cm

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Now we need to find the length of a chord that subtends 120° at the center.

This forms a triangle with:

Two sides of length 5 cm (the radii)

Unknown chord length (let's call it a)

Angle between radii = 120°

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The cosine rule relates the sides of any triangle to one of its angles:

$c^2 = a^2 + b^2 - 2ab\cos C$

Where C is the angle opposite side c.

In our case:

a = 5 cm (radius)

b = 5 cm (radius)

C = 120° (angle between radii)

c = chord length we want to find

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$\text{chord}^2 = 5^2 + 5^2 - 2(5)(5)\cos(120°)$

Key point: $\cos(120°) = -\tfrac{1}{2}$

120° is in the second quadrant where cosine is negative, and 120° = 180° - 60°, so $\cos(120°) = -\cos(60°) = -\tfrac{1}{2}$

$\text{chord}^2 = 25 + 25 - 50 \times \left(-\tfrac{1}{2}\right)$

$\text{chord}^2 = 50 + 25 = 75$

$\text{chord} = \sqrt{75}$

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$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}$

Therefore, the chord length is $5\sqrt{3}$ cm.

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Notice how the 120° chord $(5\sqrt{3} \approx 8.66 \text{ cm})$ is longer than the 60° chord $(5 \text{ cm})$? This makes sense because larger central angles create longer chords!

Answer: $5\sqrt{3}$ cm