Z twierdzenia Pitagorasa dla $△ABC$:
${\left|AB\right|}^2+{\left(\sqrt{15}\right)}^2=8^2$
${\left|AB\right|}^2+15=64$
${\left|AB\right|}^2=49|\sqrt{},|AB|>0$
$\left|AB\right|=7$

$\left|AD\right|=\left|AB\right|-\left|BD\right|=7-6=1$

$\mathrm{tg}\beta =\frac{\left|AC\right|}{\left|AD\right|}=\frac{\sqrt{15}}{1}=\sqrt{15}$