Find $\tan x$

Let $MQ=1$ . Then $HQ = 2$ and $MH = 3$ . Since $AH=HQ = 2$ . As $AB$ is a diameter of the semicircle, $\angle AMB=90^\circ$ , and $\triangle MAH$ and $\triangle MBH$ are similar. Then $\dfrac {HB}{MH} = \dfrac {MH}{AH}$ $\implies HB = \dfrac {MH}{AH} \times MH = \dfrac 32 \times 3 = \dfrac 92$ and $\tan \angle QBH = \dfrac {HQ}{HB} = \dfrac 2{\frac 92} =\dfrac 49 \approx \boxed{0.444}$

Let $|\overline {MQ}|=a\implies |\overline {HQ}|=2a$ and radius of the semicircle $=r$ . Then $|\overline {AH}|=2a,|\overline {HB}|=2r-2a$ . $\angle {AMB}$ , being a semicircular angle, $=90\degree\implies |\overline {AM}|^2+|\overline {MB}|^2=|\overline {AB}|^2\implies (4a^2+9a^2)+(9a^2+4r^2-8ra+4a^2)=4r^2\implies a=\dfrac{4r}{13}$ .

Hence $\tan x=\dfrac{|\overline {HQ}|}{|\overline {HB}|}=\dfrac {a}{r-a}=\dfrac {4}{9}\approx \boxed {0.444444}$ .