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Question #6c63a

1 Answer

Shwetank Mauria

Mar 26, 2017

You are right. It is $132^{\circ}$ $132^@$ .

Explanation:

As measure of arc $B F D = 228^{\circ}$ $BFD=228^@$ ,

$m ∠ B P D = \frac{1}{2} m ˆ B F D = \frac{1}{2} \times 228 = 114^{\circ}$ $m/_BPD=1/2m hat(BFD)=1/2xx228=114^@$

Now joining $B P$ $BP$ and $P D$ $PD$ , as $∠ C B P = ∠ C D P = 90^{\circ}$ $/_CBP=/_CDP=90^@$

$∠ B C D = 180^{\circ} - 114^{\circ} = 66^{\circ}$ $/_BCD=180^@-114^@=66^@$

and as $A E$ $AE$ subtends $∠ A O E$ $/_AOE$ at center, and ths is double $∠ B C D$ $/_BCD$

$m ˆ A F E = 2 \times 66^{\circ} = 132^{\circ}$ $mhat(AFE)=2xx66^@=132^@$

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