Determining $\sin A$ using Thales' theorem

Question

$\overline{CB}$ is a diameter of a circle with a radius of 2 cm and center $O$. $\triangle ABC$ is a right triangle, and $\overline{CD}$ has the length $\sqrt 3$. Determine $\sin A$ (Hint: Use Thales’ Theorem).

Let's take a look at Thales's theorem

Thales' theorem states that if $A$, $B$, and $C$ are distinct points on a circle where the line $\overline{AC}$ is a diameter, then the angle $\angle ABC $ is a right angle.

Thereby, we have that $\angle D = 90^\circ$, which also yields $\triangle ACD$ is a right triangle. However, I couldn't proceed.

Answer 1

1)$\angle DCB = \angle A$ (Why?)

2)$ \triangle BDC$ is a right triangle, Thales circle over $CB$.

$|BD|^2= 4^2 -(√3)^2= 13$ (Pythagoras)

$\sin A = \sin (\angle DCB) = \sqrt{13}/4.$

Answer 2

You know that $\cos A=\frac{\,\overline{AD}\,}{\,\overline{AC}\,}$. Since the triangles $\triangle ADC$ and $\triangle CDB$ are similar (this is a consequence of Thales' theorem),$$\cos A=\frac{\,\overline{AD}\,}{\,\overline{AC}\,}=\frac{\,\overline{CD}\,}{\,\overline{CB}\,}=\frac{\sqrt3}4$$and therefore $\sin A=\frac{\sqrt{13}}4$.