How to find angle in circle?

Question

I have done question $(a)$ but cannot do question $(b)$.

Does anyone have any solutions or ideas?

Thank you.

Answer 1

Let $M$ be the midpoint of $BC$. $\angle CAM=\alpha/4$.

$$\sin\frac{\alpha}{4}=\frac{BC/2}{AC}=\frac{r/2}{2r}=\frac{1}{4}$$

Answer 2

HINT: for a) use that $$A=\frac{r^2}{2}(x-\sin(x))$$

Answer 3

Consider $\triangle ABC$. By cosine rule,

$$\begin{align*} BC^2 &= AB^2 + AC^2 - 2 AB\cdot AC\cos\angle BAC\\ r^2 &= 4r^2 + 4r^2 - 8r^2\cos\angle BAC\\ 8\cos \angle BAC &= 7\\ \cos\angle BAC &= \frac 78\\ 1 - 2\sin^2\frac{\angle BAC}2 &= \frac 78\\ \sin^2\frac{\angle BAC}2 &= \frac1{16}\\ \sin\frac{\angle BAC}2 &= \frac14 \quad\text{(positive square root taken)}\\ \angle BAC &= 2\arcsin \frac14\\ \alpha &= 4\arcsin \frac14 \end{align*}$$