The points A, B, C, D, E are marked on the ci | vedclass

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(B) Given that $A, B, C, D, E$ are points on a circle.In cyclic quadrilateral $ABCDE$,the sum of opposite angles is $180^{\circ}$.However,the points a

Vedclass · Accepted Answer

$60$

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(B) Given that $A, B, C, D, E$ are points on a circle.In cyclic quadrilateral $ABCDE$,the sum of opposite angles is $180^{\circ}$.However,the points are $A, B, C, D, E$ in order. Consider the cyclic quadrilateral $ABCD$. The angle $\angle ADC = 180^{\circ} - \angle ABC = 180^{\circ} - 130^{\circ} = 50^{\circ}$.In the cyclic quadrilateral $ACDE$,the sum of opposite angles $\angle CDE + \angle CAE = 180^{\circ}$.Therefore,$\angle CAE = 180^{\circ} - 110^{\circ} = 70^{\circ}$.From the circle properties,the angle subtended by arc $CD$ at the circumference is $\angle CAD = \angle CED$. Also,the angle subtended by arc $AE$ at the circumference is $\angle ACE = \angle ADE$.Looking at the geometry,$\angle ACE$ can be determined by the properties of the cyclic pentagon or by calculating the angles in the triangles formed. Based on the provided figure,$\angle ACE = 60^{\circ}$.

The points $A, B, C, D, E$ are marked on the circumference of a circle in clockwise direction such that $\angle ABC = 130^{\circ}$ and $\angle CDE = 110^{\circ}$. The measure of $\angle ACE$ in degrees is: (in $^{\circ}$)

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