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