Can someone help me solve for this angle?

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1 Answer
CW
Apr 28, 2018

angleAFD=35^@∠AFD=35∘

Explanation:

enter image source here
Let OO be the center of the circle.
angleCAG=180-65-80=35^@∠CAG=180−65−80=35∘
angleOCA=90-angleACG=90-65=25^@∠OCA=90−∠ACG=90−65=25∘
As OA=OC, => angleOAC=angleOCA=25^@OA=OC,⇒∠OAC=∠OCA=25∘
=> angleFAB=90-angleOAC-angleCAG=90-25-35=30^@⇒∠FAB=90−∠OAC−∠CAG=90−25−35=30∘
given angleDOC=100^@∠DOC=100∘,
as OD=OC, => angleODC=angleOCD=(180-100)/2=40^@OD=OC,⇒∠ODC=∠OCD=180−1002=40∘
=> angleDCA=angleOCD+angleOCA=40+25=65^@⇒∠DCA=∠OCD+∠OCA=40+25=65∘
As ABCDABCD is a cyclic quadrilateral,
=> angleDBA=angleDCA=65^@⇒∠DBA=∠DCA=65∘
=> angleABF=180-65=115^@⇒∠ABF=180−65=115∘
Now consider DeltaABF,
=> angleAFB=angleAFD=x=180-30-115=35^@

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