Question : ABCD is a cyclic quadrilateral. Sides AB and DC, when produced, meet at E and sides AD and BC, meet at F. If $\angle $ADC = 76$^\circ$ and $\angle$AED = 55$^\circ$, then $\angle$AFB is equal to:
Option 1: $34^\circ$
Option 2: $26^\circ$
Option 3: $29^\circ$
Option 4: $27^\circ$
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Correct Answer: $27^\circ$
Solution :
We know that the sum of all the sides of the triangle is $180^\circ $. In $\triangle DAE$, $⇒ \angle ADC + \angle AED + \angle DAE = 180^\circ$ $⇒ 76^\circ+ 55^\circ+ \angle DAE = 180^\circ$ $⇒ 131^\circ+ \angle DAE = 180^\circ$ $⇒ \angle DAE = (180^\circ - 131^\circ)$ $⇒ \angle DAE = 49^\circ$ The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. $⇒ \angle DAE = \angle DCF = 49^\circ $ Now, An exterior angle of a triangle is equal to the sum of the two interior opposite angles. $⇒ \angle ADC = \angle DCF + \angle AFB$ $⇒ 76^\circ= 49^\circ+ \angle AFB$ $⇒ \angle AFB = (76^\circ - 49^\circ)$ $\therefore \angle AFB = 27^\circ$ Hence, the correct answer is $27^\circ$.
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