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Question Number 67774 by TawaTawa last updated on 31/Aug/19

Answered by MJS last updated on 31/Aug/19

$$DE=AB=2r$$$$\mathrm{area}\mathrm{of}\mathrm{semicircle}=\frac{\pi }{2}{r}^2$$$$\mathrm{area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\mid AB\mid h=rh$$$$\mathrm{with}h=r\tan x$$$$=r^2\tan x$$$$\Rightarrow$$$$\frac{\pi }{2}{r}^2=r^2\tan x$$$$\Rightarrow x=\arctan \frac{\pi }{2}\approx 57.52°$$

Commented by TawaTawa last updated on 31/Aug/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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