In the given figure,if angle BCD = 40^{circ} | vedclass

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(C) $1$. In cyclic quadrilateral $ABDE$,the exterior angle is equal to the interior opposite angle. Thus,$\angle CDE = \angle BAE = 65^{\circ}$. Since

Vedclass · Accepted Answer

$a=150^{\circ}, b=10^{\circ}, c=65^{\circ}, d=75^{\circ}$

Vedclass · Answer

(C) $1$. In cyclic quadrilateral $ABDE$,the exterior angle is equal to the interior opposite angle. Thus,$\angle CDE = \angle BAE = 65^{\circ}$. Since $CDE$ is a straight line,$\angle c = 180^{\circ} - 65^{\circ} = 115^{\circ}$ is incorrect based on the diagram geometry. Let's re-evaluate: In $	riangle ACE$,$\angle C = 40^{\circ}$ and $\angle CAE = 65^{\circ}$. Thus,$\angle AEC = 180^{\circ} - (40^{\circ} + 65^{\circ}) = 75^{\circ}$.$2$. Since $ABDE$ is a cyclic quadrilateral,the exterior angle $\angle CDE = \angle BAE = 65^{\circ}$. Therefore,$c = 180^{\circ} - 65^{circ} = 115^{\circ}$ is not correct as $c$ is an interior angle of the cyclic quadrilateral. Actually,$\angle BDE = 180^{\circ} - 65^{\circ} = 115^{\circ}$.$3$. From the cyclic property,$\angle BDE + \angle BAE = 180^{\circ} \implies \angle BDE = 180^{\circ} - 65^{\circ} = 115^{\circ}$.$4$. $\angle d = \angle AED = 180^{\circ} - 75^{\circ} = 105^{\circ}$ is wrong. Let's use $	riangle ACD$: $\angle CAD = 65^{\circ}, \angle ACD = 40^{\circ} \implies \angle ADC = 180^{\circ} - 105^{\circ} = 75^{\circ}$.$5$. Since $ABDE$ is cyclic,$\angle BDE = 180^{\circ} - 65^{\circ} = 115^{\circ}$. Thus,$c = 180^{\circ} - 115^{\circ} = 65^{\circ}$.$6$. $\angle a = 180^{\circ} - \angle BDE = 180^{\circ} - 30^{\circ} = 150^{\circ}$.$7$. In $	riangle FDE$,$\angle FDE = 180^{\circ} - 115^{\circ} = 65^{\circ}$ and $\angle FED = 180^{\circ} - 105^{\circ} = 75^{\circ}$. Thus,$b = 180^{\circ} - (65^{\circ} + 105^{\circ}) = 10^{\circ}$.$8$. Therefore,$a=150^{\circ}, b=10^{\circ}, c=65^{\circ}, d=75^{\circ}$.

In the given figure,if $\angle BCD = 40^{\circ}$ and $\angle BAE = 65^{\circ},$ then find the values of $a, b,$ $c$ and $d$.

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