find the tan

Let the center of the semicircle be $O$ and $d=6$ . Then $\triangle OST$ is a right triangle at $T$ . Since $OS=5$ and $OT=3$ , then $ST=4$ . Let $MN$ and $TR$ be perpendicular to $AS$ , then $\triangle OST$ and $\triangle RST$ are similar. Then $TR = 2.4$ and $RS=3.2$ . And $AR = AS-RS = 8 - 3.2= 4.8$ . Again $\triangle AMN$ and $\triangle ATR$ are similar are $M$ is the midpoint of $AT$ , then $MN = \frac 12 TR = 1.2$ and $AN = \frac 12 AR = 2.4$ . Then

$\begin{aligned} \tan \angle MSA & = \frac {MN}{NS} = \frac {MN}{AS - AN} = \frac {1.2}{8-2.4} = \frac {1.2}{5.6} = \frac 3{14} \approx \boxed{0.214} \end{aligned}$

Let $\angle {TAS}=α$ . Then $|\overline {AM}|=\dfrac{d}{2}\cos α$ .

Let the center of the semicircle be $O$ . Then $\angle {STO}=90\degree\implies \cos (2α)=\frac{3}{5},\sin (2α)=\frac{4}{5}$ .

Applying sine rule to $\triangle {AMS}$ we get

$\dfrac{|\overline {AM}|}{\sin \theta}=\dfrac{|\overline {AS}|}{\sin (α+\theta) }\implies$

$8\sin \theta=3\sin (α+\theta) \cos α\implies 13\sin \theta=3\sin (2α+\theta) =3\sin (2α)\cos \theta+3\cos (2α)\sin \theta=\frac{12}{5}\cos \theta+\frac{9}{5}\sin \theta\implies \tan \theta=\dfrac{3}{14}\approx \boxed {0.21428}$ .