Question

If the given figure, $AB,BC$ and $CD$ are equal chords of a circle with centre $O$ and $AD$ is a diameter. If $\angle DEF={110}^o$. Then (i) $\angle AEF$ (ii) $\angle FAB$ are respectively:

• A. ${20}^o\&{130}^o$
• B. ${30}^o\&{130}^o$
• C. ${20}^o\&{120}^o$
• D. ${15}^o\&{130}^o$

Answer 1

The correct option is A ${20}^o\&{130}^o$

$Given-ABCDEisapolygoninccribedinacirclewithcentreOandthediameterasAD.AFhasbeenjoined.\angle FED={110}^o.AB=BC=CD.Tofindout-\left(i\right)\angle AEF=?\left(ii\right)\angle FAB=?Solution-WejoinOB\&OC.Between\Delta OAB\&\Delta OBCOA=OB=OC(radiiofthesamecircleandAB=BC).\therefore BySSStest,\Delta OAB\cong \Delta OBC⟹OA=BC.But,BC=AB.\therefore OA=AB=BC.So\Delta OABisanequilateralone.i.e\angle OAB=\angle ABO=\angle AOB={60}^o.(Eachangleofanequilateraltriangle={60}^o).Again,AFEDisacyclicquadrilateral.\therefore \angle FAD+\angle FED={180}^o⟹\angle FAD={180}^o-\angle FED={180}^o-{110}^o.={70}^o.So,\angle FAB=\angle FAD+\angle OAB={70}^o+{60}^o={130}^o.........\left(ansii\right).Now,\angle AED={90}^o\left(angleinasemicircle\right).\angle AEF=\angle FED-\angle AED=\therefore \angle AEF={110}^o-{90}^o={20}^o........\left(ansi\right)So,\angle AEF\&\angle FABarerespectively{20}^o\&{130}^o.Hence,optionCiscorrect.$