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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:
243842_4e306e0141354bb491c92f90a351437c.png


A
20o&130o
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B
30o&130o
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C
20o&120o
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D
15o&130o
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Solution

The correct option is A $${ 20 }^{ o }\quad \& \quad { 130 }^{ o }$$
$$ Given-\\ ABCDE\quad is\quad a\quad polygon\quad inccribed\quad in\quad a\quad circle\quad with\quad centre\quad O\\ and\quad the\quad diameter\quad as\quad AD.\\ AF\quad has\quad been\quad joined.\\ \angle FED={ 110 }^{ o }.\\ AB=BC=CD.\\ To\quad find\quad out-\\ (i)\quad \angle AEF=?\quad \quad (ii)\quad \angle FAB=?\\ Solution-\\ We\quad join\quad OB\quad \& \quad OC.\\ Between\quad \Delta OAB\quad \& \quad \Delta OBC\\ OA=OB=OC(radii\quad of\quad the\quad same\quad circle\quad and\\ AB=BC).\\ \therefore \quad By\quad SSS\quad test,\quad \Delta OAB\cong \Delta OBC\Longrightarrow OA=BC.\\ But,\quad BC=AB.\\ \therefore OA=AB=BC.\\ So\quad \Delta OAB\quad is\quad an\quad equilateral\quad one.\\ i.e\quad \angle OAB=\angle ABO=\angle AOB={ 60 }^{ o }. \left( Each\quad angle\quad of\quad an\quad equilateral\quad triangle={ 60 }^{ o } \right) .\\ Again,\quad AFED\quad is\quad a\quad cyclic\quad quadrilateral.\\ \therefore \quad \angle FAD+\angle FED={ 180 }^{ o }\Longrightarrow \angle FAD={ 180 }^{ o }-\angle FED\\ ={ 180 }^{ o }-{ 110 }^{ o }.={ 70 }^{ o }.\\ So,\quad \angle FAB=\angle FAD+\angle OAB\\ ={ 70 }^{ o }+60^{ o }={ 130 }^{ o }.........(ans\quad ii).\\ Now,\quad \angle AED={ 90 }^{ o }(angle\quad in\quad a\quad semicircle).\\ \angle AEF=\angle FED-\angle AED=\\ \therefore \quad \angle AEF={ 110 }^{ o }-{ 90 }^{ o }={ 20 }^{ o }........(ans\quad i)\\ So,\quad \angle AEF\quad \& \quad \angle FAB\quad are\quad respectively\quad { 20 }^{ o }\quad \& \quad { 130 }^{ o }.\\ Hence,\quad option\quad C \quad is \quad correct.$$
305348_243842_ans_a62d55a33e3340129a5b9326b5f26a49.png

Mathematics

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