In a cyclic quadrilateral ABCD,angle A : angl | vedclass

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Question

(D) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Given $\angle A : \angle C = 4 : 5$,let $\angle A = 4x$ and $\angle C = 5x$.

Vedclass · Accepted Answer

$\angle A = 80^{\circ}, \angle B = 75^{\circ}, \angle C = 100^{\circ}, \angle D = 105^{\circ}$

Vedclass · Answer

(D) In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.Given $\angle A : \angle C = 4 : 5$,let $\angle A = 4x$ and $\angle C = 5x$.Since $\angle A + \angle C = 180^{\circ}$,we have $4x + 5x = 180^{\circ} \Rightarrow 9x = 180^{\circ} \Rightarrow x = 20^{\circ}$.Thus,$\angle A = 4(20^{\circ}) = 80^{\circ}$ and $\angle C = 5(20^{\circ}) = 100^{\circ}$.Given $\angle B : \angle D = 5 : 7$,let $\angle B = 5y$ and $\angle D = 7y$.Since $\angle B + \angle D = 180^{\circ}$,we have $5y + 7y = 180^{\circ} \Rightarrow 12y = 180^{\circ} \Rightarrow y = 15^{\circ}$.Thus,$\angle B = 5(15^{\circ}) = 75^{\circ}$ and $\angle D = 7(15^{\circ}) = 105^{\circ}$.Therefore,the angles are $\angle A = 80^{\circ}, \angle B = 75^{\circ}, \angle C = 100^{\circ}, \angle D = 105^{\circ}$.

In a cyclic quadrilateral $ABCD$,$\angle A : \angle C = 4 : 5$ and $\angle B : \angle D = 5 : 7$. Find all the angles of $ABCD$.

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