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Question Number 112076 by I want to learn more last updated on 06/Sep/20

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

$$\mathrm{Is}\mathrm{OD}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\mathrm{D}?$$

Commented by som(math1967) last updated on 06/Sep/20

$$\mathrm{yes}\mathrm{you}\mathrm{are}\mathrm{correct}$$

Commented by Aina Samuel Temidayo last updated on 06/Sep/20

$$\mathrm{DC}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{chord}.\mathrm{I}\mathrm{think}\mathrm{your}$$$$\mathrm{solution}\mathrm{is}\mathrm{incorrect}.$$

Answered by mr W last updated on 06/Sep/20

$$sayradiusOC=OB=1$$$$BC=\frac{2}{\cos 20°}$$$$\frac{CD}{\sin 30°}=\frac{BD}{\sin 40°}=\frac{BC}{\sin 70°}=\frac{2}{{\cos }^{2}20°}$$$$\Rightarrow CD=\frac{4}{{\cos }^{2}20°}$$$${OD}^2=1^2+\frac{4^2}{{\cos }^{4}20°}-2\times \frac{4}{{\cos }^{2}20°}\times \cos 40°$$$$=1+\frac{16}{{\cos }^{4}20°}-\frac{8(2{\cos }^{2}20-1)}{{\cos }^{2}20°}$$$$=\frac{16}{{\cos }^{4}20°}+\frac{8}{{\cos }^{2}20°}-15$$$$=\frac{1}{{\cos }^{4}20°}(16+8{\cos }^{2}20°-15{\cos }^{4}20°)$$$$\frac{\sin x}{OC}=\frac{\sin 20°}{OD}$$$$\Rightarrow \sin x=\frac{\sin 20°{\cos }^{2}20°}{\sqrt{16+8{\cos }^{2}20°-15{\cos }^{4}20°}}$$$$\Rightarrow x\approx 5.139°$$

Commented by I want to learn more last updated on 06/Sep/20

$$\mathrm{Wow},\mathrm{thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$$$\mathrm{Sir},\mathrm{what}\mathrm{condition}\mathrm{can}\mathrm{i}\mathrm{say}\mathrm{radii}\mathrm{OC}=\mathrm{OB}=1??$$

Commented by mr W last updated on 06/Sep/20

$$youcansayOC=OB=r,it{doesn}^{\prime }t$$$$matter,sincewearecalculatingthe$$$$anglesonly.$$

Commented by I want to learn more last updated on 06/Sep/20

$$\mathrm{Ohh},\mathrm{i}\mathrm{understand}\mathrm{sir}.\mathrm{Thanks}\mathrm{so}\mathrm{much}.$$

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