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Question #6c63a

1 Answer

Shwetank Mauria

Mar 26, 2017

You are right. It is $132^@$ .

Explanation:

As measure of arc $BFD=228^@$ ,

$m/_BPD=1/2m hat(BFD)=1/2xx228=114^@$

Now joining $BP$ and $PD$ , as $/_CBP=/_CDP=90^@$

$/_BCD=180^@-114^@=66^@$

and as $AE$ subtends $/_AOE$ at center, and ths is double $/_BCD$

$mhat(AFE)=2xx66^@=132^@$

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