Angle to a point on a chord

Question

Calculate angle $\alpha$ in the figure given the circle's radius $R$ and the distances $d=\overline{CM}$ and $\overline{CD}$.

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Answer 1

Established the equation below with the sine sule applied to $\triangle ACM$,

$$\frac{\sin \alpha}{\sin (45+\alpha)}=\frac dR\implies \tan\alpha =\frac d{\sqrt2 R-d}$$

which yields $\alpha = \tan^{-1}\frac d{\sqrt2 R-d}$.

Answer 2

$ACD$ is an isosceles right triangle.

Consider triangle $CMA$ to apply Sine Rule:

$$ \dfrac{CM=d}{R}=\dfrac{\sin \alpha}{\sin (\pi/4+\alpha)}=\dfrac{\sqrt 2 \sin \alpha}{\sin \alpha + \cos \alpha};$$ $$\tan\alpha =\dfrac d{\sqrt2 R-d};$$