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Question Number 139506 by I want to learn more last updated on 28/Apr/21

Commented by I want to learn more last updated on 28/Apr/21

$$\mathrm{Find}\mathrm{the}\mathrm{double}\mathrm{line}\mathrm{angle}.$$

Commented by som(math1967) last updated on 28/Apr/21

$$60°$$

Commented by I want to learn more last updated on 28/Apr/21

$$\mathrm{workings}\mathrm{sir}.$$

Commented by som(math1967) last updated on 28/Apr/21

Commented by som(math1967) last updated on 28/Apr/21

$$let\mathrm{\angle }AOB=x$$$$\therefore \mathrm{\angle }ABC=x$$$$[AO\parallel BC,AB\parallel OC$$$$ABCOisparallelogram$$$$\therefore \mathrm{\angle }AOB=\mathrm{\angle }ABC]$$$$\mathrm{\angle }AOC=2\mathrm{\angle }ADC$$$$\Rightarrow \mathrm{\angle }ADC=\frac{1}{2}\mathrm{\angle }AOB=\frac{x}{2}$$$$ABCDCyclic$$$$\therefore \mathrm{\angle }ABC+\mathrm{\angle }ADC=180$$$$x+\frac{x}{2}=180$$$$x=120°$$$$\mathrm{\angle }OCB=180-120=60°$$$$[AO\parallel BC]$$

Commented by I want to learn more last updated on 28/Apr/21

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

Answered by mr W last updated on 28/Apr/21

Commented by mr W last updated on 28/Apr/21

$$OA=OB=OC=radius$$$$AB=OC=OA=OB$$$$\Rightarrow \mathrm{\Delta }OABisequilateraltriangle$$$$\Rightarrow \mathrm{\angle }OAB=60°$$

Commented by I want to learn more last updated on 28/Apr/21

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.\mathrm{Thanks}.$$

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