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
$$\mathrm{60}° \\ $$
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 ∠AOB=x ∴∠ABC=x [AO∥BC ,AB∥OC ABCO is parallelogram ∴∠AOB=∠ABC] ∠AOC=2∠ADC ⇒∠ADC=(1/2)∠AOB=(x/2) ABCD Cyclic ∴∠ABC+∠ADC=180 x+(x/2)=180 x=120° ∠OCB=180−120=60° [AO∥BC]](Q139514.png)
$${let}\:\angle{AOB}={x} \\ $$$$\therefore\angle{ABC}={x} \\ $$$$\left[{AO}\parallel{BC}\:,{AB}\parallel{OC}\right. \\ $$$${ABCO}\:{is}\:{parallelogram} \\ $$$$\left.\therefore\angle{AOB}=\angle{ABC}\right] \\ $$$$\angle{AOC}=\mathrm{2}\angle{ADC} \\ $$$$\Rightarrow\angle{ADC}=\frac{\mathrm{1}}{\mathrm{2}}\angle{AOB}=\frac{{x}}{\mathrm{2}} \\ $$$${ABCD}\:{Cyclic} \\ $$$$\therefore\angle{ABC}+\angle{ADC}=\mathrm{180} \\ $$$${x}+\frac{{x}}{\mathrm{2}}=\mathrm{180} \\ $$$${x}=\mathrm{120}° \\ $$$$\angle{OCB}=\mathrm{180}−\mathrm{120}=\mathrm{60}° \\ $$$$\left[{AO}\parallel{BC}\right] \\ $$
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
$${OA}={OB}={OC}={radius} \\ $$$${AB}={OC}={OA}={OB} \\ $$$$\Rightarrow\Delta{OAB}\:{is}\:{equilateral}\:{triangle} \\ $$$$\Rightarrow\angle{OAB}=\mathrm{60}° \\ $$
Commented by I want to learn more last updated on 28/Apr/21
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{Thanks}. \\ $$