Q. 305.0( 1 Vote )

# In the given figu

Given: CD||AB and BAD = 30°

Consider ΔABD

ADB = 90° (angle in semicircle)

Now, by angle sum property

ABD + 30° + 90° = 180°

ABD = 180° – 30° – 90°

ABD = 60°

Here,

ABD + ACD = 180° (opposite angles in cyclic quadrilateral are supplementary)

60° + ACD = 180°

BCD = 180° – 60°

BCD = 120°

Here, CD||AB and AC is the transversal

CAB + ACD = 180° (interior angles along the transversal are supplementary)

CAB + 120° = 180°

ABC = 180° – 120° = 60°

ABC = 60°

CAD = 60° – 30° = 30°

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