Question Number 67774 by TawaTawa last updated on 31/Aug/19
Answered by MJS last updated on 31/Aug/19
$${DE}={AB}=\mathrm{2}{r} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{semicircle}\:=\frac{\pi}{\mathrm{2}}{r}^{\mathrm{2}} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:=\frac{\mathrm{1}}{\mathrm{2}}\mid{AB}\mid{h}={rh} \\ $$$$\:\:\:\:\:\mathrm{with}\:{h}={r}\mathrm{tan}\:{x} \\ $$$$\:\:\:\:\:={r}^{\mathrm{2}} \mathrm{tan}\:{x} \\ $$$$\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}{r}^{\mathrm{2}} ={r}^{\mathrm{2}} \mathrm{tan}\:{x} \\ $$$$\Rightarrow\:{x}=\mathrm{arctan}\:\frac{\pi}{\mathrm{2}}\:\approx\mathrm{57}.\mathrm{52}° \\ $$
Commented by TawaTawa last updated on 31/Aug/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$