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Question Number 7945 by Rasheed Soomro last updated on 25/Sep/16

Commented by Rasheed Soomro last updated on 25/Sep/16

$$InabovefigureAF,adiameterofthe$$$$circlewithcentreGhasbeentrisected$$$$atB\mathrm{\&}D(i-eAB=BD=DF)$$$$IsthereanypointHonthecirclefor$$$$which$$$$\mathrm{\angle }AHB=\mathrm{\angle }BHD=\mathrm{\angle }DHF$$$$I-\mathcal{E}\beta =\gamma =\delta ?$$$$Ifyesthenwhatis\mathrm{\angle }HGF?$$

Commented by prakash jain last updated on 27/Sep/16

$$\mathrm{\angle }AHF=\frac{\pi }{2}=90°$$$$\mathrm{if}\mathrm{\angle }\mathrm{AHB}=\mathrm{\angle }\mathrm{BHD}=\mathrm{\angle }\mathrm{DHF}$$$$\mathrm{then}\mathrm{\angle }\mathrm{AHB}=\mathrm{\angle }\mathrm{BHD}=\mathrm{\angle }\mathrm{DHF}=\frac{\pi }{6}=30°$$$$$$$$\mathrm{BG}=r-\frac{2r}{3}=\frac{r}{3}=\mathrm{GD}$$$$\mathrm{Let}\mathrm{us}\mathrm{say}\mathrm{\angle }\mathrm{BHG}=x\Rightarrow \mathrm{\angle }\mathrm{GHD}=\frac{\pi }{6}-x$$$$△\mathrm{HBG},\mathrm{\angle }\mathrm{HGB}=\pi -\alpha$$$$\mathrm{\angle }\mathrm{HBG}=\pi -x-(\pi -\alpha )=\alpha -x$$$$\frac{\mathrm{HG}}{\sin (\alpha -x)}=\frac{\mathrm{BG}}{\sin x}$$$$△\mathrm{GHD}$$$$\mathrm{\angle }\mathrm{HGD}=\alpha$$$$\mathrm{\angle }\mathrm{GHD}=\frac{\pi }{6}-x$$$$\mathrm{\angle }\mathrm{HDG}=\pi -\alpha -(\frac{\pi }{6}-x)=\frac{5\pi }{6}+x-\alpha$$$$\frac{\mathrm{HG}}{\sin (\frac{5\pi }{6}+x-\alpha )}=\frac{\mathrm{GD}}{\sin (\frac{\pi }{6}-x)}=\frac{\mathrm{BG}}{\sin (\frac{\pi }{6}-x)}$$$$\frac{\sin x}{\mid \sin (\alpha -x)\mid }=\frac{\sin (\frac{\pi }{6}-x)}{\mid \sin (\frac{5\pi }{6}+x-\alpha )\mid }$$$$△\mathrm{HAD}$$$$\mathrm{\angle }\mathrm{HAD}=\pi -\frac{\pi }{3}-(\frac{5\pi }{6}+x-\alpha )=\alpha -x-\frac{\pi }{6}$$$$△\mathrm{HAB}$$$$\frac{\mathrm{HB}}{\sin (\alpha -x-\frac{\pi }{6})}=\frac{\mathrm{AB}}{\sin \frac{\pi }{6}}$$$$△\mathrm{HBD}$$$$\frac{\mathrm{BD}}{\sin \frac{\pi }{6}}=\frac{\mathrm{HB}}{\sin (\frac{5\pi }{6}+x-\alpha )}$$$$\mathrm{AB}=\mathrm{BD}\Rightarrow \alpha -x-\frac{\pi }{6}=\frac{5\pi }{6}+x-\alpha$$$$2\alpha -2x=\pi \Rightarrow \alpha =\frac{\pi }{2}+x$$$$\frac{\sin x}{\mid \sin (\frac{\pi }{2}+x-x)\mid }=\frac{\sin (\frac{\pi }{6}-x)}{\mid \sin (\frac{5\pi }{6}-\frac{\pi }{2})\mid }$$$$\sin x\cdot \mid \sin \left(\frac{\pi }{3}\right)\mid =\sin (\frac{\pi }{6}-x)$$$$\frac{\sqrt{3}}{2}\sin x=\sin (\frac{\pi }{6}-x)$$$$\frac{\sqrt{3}}{2}\sin x=\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x$$$$\tan x=2\sqrt{3}$$$$x={\tan }^{-1}2\sqrt{3}$$$$\alpha =\frac{\pi }{2}+{\tan }^{-1}2\sqrt{3}$$$$\mathrm{Please}\mathrm{check}.$$

Commented by Rasheed Soomro last updated on 27/Sep/16

$$\mathcal{GREAT}approach!!!$$$$I{didn}^{\prime }tunderstandsomelines:$$$$△\mathrm{GHD}$$$$⋮$$$$\mathrm{\angle }\mathrm{HDG}=\pi -\alpha -(\frac{\pi }{6}-x)=\frac{5\pi }{6}+x-\alpha$$$$\frac{\mathrm{HG}}{\sin \left(\frac{5\pi }{6}\stackrel{?}{+}\alpha \stackrel{?}{-}x\right)}=\frac{\mathrm{GD}}{\sin (\frac{\pi }{6}-x)}=\frac{\mathrm{BG}}{\sin (\frac{\pi }{6}-x)}$$$$.....$$$$...$$

Commented by prakash jain last updated on 27/Sep/16

$$\mathrm{Yes}.\mathrm{My}\mathrm{mistake}.\mathrm{I}\mathrm{will}\mathrm{correct}.$$

Answered by prakash jain last updated on 02/Oct/16

$$\mathrm{In}\mathrm{comments}.$$

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