The figure below shows a triangle inside a circle. AB = 8 cm, BC = 10 cm and AC = 13 cm. a) Calculate the area of triangle ABC. b) Angle BAC

The side lengths are $a = BC = 10$ cm, $b = AC = 13$ cm, and $c = AB = 8$ cm.

$$S = \frac{1}{2}(a+b+c) = \frac{1}{2}(10 + 13 + 8) = \frac{1}{2}(31) = 15.5 \text{ cm}$$

$$\text{Area} = \sqrt{S(S-a)(S-b)(S-c)}$$ $$\text{Area} = \sqrt{15.5(15.5-10)(15.5-13)(15.5-8)}$$ $$\text{Area} = \sqrt{15.5(5.5)(2.5)(7.5)}$$ $$\text{Area} = \sqrt{1598.4375}$$ $$\text{Area} \approx 39.98046 \text{ cm}^2$$

Rounding to two decimal places:
The area of triangle ABC is $\boxed{39.98 \text{ cm}^2}$.

Let $\theta = \angle BAC$. The sides adjacent to $\angle BAC$ are $AB=c=8$ cm and $AC=b=13$ cm.

$$\text{Area} = \frac{1}{2}bc \sin \theta$$ $$39.98046 = \frac{1}{2} \times 13 \times 8 \times \sin \theta$$ $$39.98046 = 52 \sin \theta$$

$$\sin \theta = \frac{39.98046}{52}$$ $$\sin \theta \approx 0.768855$$ $$\theta = \arcsin(0.768855)$$ $$\theta \approx 50.24^\circ$$

Rounding to two decimal places:
Angle BAC is $\boxed{50.24^\circ}$.