Isoceles Triangle finding angle problem

Question

In the diagram below $AB=AC=BD=DE$ and $\angle BAC=30^o$, what is the value of $\angle CED$ in degree?

Answer 1

Let $AB=BD=DE=AC=x$ and $\angle CED=\alpha$. In rght angled triangle AFC we have:

$$CF=x \sin(30)=\frac x 2$$

$$\Rightarrow BC=\frac{CF}{ \sin(75)}=\frac x{ 2\sin(75)}$$

$\Rightarrow CD=BD-BC=x(1-\frac1{2\sin(75)})$

In triangle ECD due to sine rule we can write:

$$\frac x{\sin(105)}=\frac{x(1-\frac1{2\sin(75)})}{\sin(\alpha)}$$

$$\sin(\alpha)=\Big(1-\frac1{2\sin(75)}\Big)\sin(105)$$

$\sin(105)=\sin(180-75)=\sin(75)$

$\Rightarrow \sin(\alpha)=\frac 12[2\sin(75)-1]\approx 0.466$

Which gives $ \alpha\approx 27.8^o$