To solve the problem given in the figure:

Given Information:

1. The radius of the circle is $5 \, \text{cm}$.
2. $\angle BOA = 120^\circ$, where $O$ is the center of the circle.

Part (a): Find $\angle C$

From the diagram:

• $\triangle AOC$ is an isosceles triangle (OA = OC as both are radii of the circle).
• $\angle BOA = 120^\circ$.

The angle subtended by arc $BA$ at point $C$ on the circumference of the circle is half of $\angle BOA$ (inscribed angle theorem). Thus: $\angle C = \frac{\angle BOA}{2} = \frac{120^\circ}{2} = 60^\circ.$

$\angle C = 60^\circ$.

Part (b): Find the length of AC

$\triangle AOC$ is isosceles with:

• OA = OC = $5 \, \text{cm}$.
• $\angle AOC = \angle BOA = 120^\circ$.

Using the cosine rule in $\triangle AOC$: $AC^2 = OA^2 + OC^2 - 2 \cdot OA \cdot OC \cdot \cos(\angle AOC).$ Substitute the values: $AC^2 = 5^2 + 5^2 - 2 \cdot 5 \cdot 5 \cdot \cos(120^\circ).$ $\cos(120^\circ) = -\frac{1}{2}$, so: $AC^2 = 25 + 25 - 2 \cdot 5 \cdot 5 \cdot \left(-\frac{1}{2}\right),$ $AC^2 = 25 + 25 + 25 = 75.$ $AC = \sqrt{75} = 5\sqrt{3} \, \text{cm}.$

Length of AC = $5\sqrt{3} \, \text{cm}$.

Final Answers:

1. $\angle C = 60^\circ$,
2. Length of $AC = 5\sqrt{3} \, \text{cm}$.

Would you like further clarification or a more detailed explanation?

Related Questions:

1. What is the derivation of the cosine rule used here?
2. How is the inscribed angle theorem applied in similar problems?
3. How would the calculations change if the angle at $BOA$ were different?
4. What is the area of $\triangle AOC$?
5. How do you calculate the perimeter of $\triangle AOC$?

Tip: Always sketch diagrams to confirm relationships between angles and sides before applying trigonometric formulas.