In a cyclic quadrilateral ABCD,angle A = 4 an | vedclass

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(B) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 180

Vedclass · Accepted Answer

$\angle A = 144^{\circ}, \angle B = 135^{\circ}, \angle C = 36^{\circ}, \angle D = 45^{\circ}$

Vedclass · Answer

(B) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Therefore,$\angle A + \angle C = 180^{\circ}$ and $\angle B + \angle D = 180^{\circ}$.Given $\angle A = 4 \angle C$,substituting this into the first equation: $4 \angle C + \angle C = 180^{\circ} \implies 5 \angle C = 180^{\circ} \implies \angle C = 36^{\circ}$.Then,$\angle A = 4 	imes 36^{\circ} = 144^{\circ}$.Given $\angle B = 3 \angle D$,substituting this into the second equation: $3 \angle D + \angle D = 180^{\circ} \implies 4 \angle D = 180^{\circ} \implies \angle D = 45^{\circ}$.Then,$\angle B = 3 	imes 45^{\circ} = 135^{\circ}$.Thus,the angles are $\angle A = 144^{\circ}, \angle B = 135^{\circ}, \angle C = 36^{\circ}, \angle D = 45^{\circ}$.

In a cyclic quadrilateral $ABCD$,$\angle A = 4 \angle C$ and $\angle B = 3 \angle D$. Find all the angles of $ABCD$.

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