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In the given figure, AOB is a diameter ...

In the given figure, `AOB` is a diameter of a circle and `CD||AB` . If `/_ BAD = 30^(@)` then `/_ CAD = ?`

A

`30^(@)`

B

`60^(@)`

C

`45^(@)`

D

`50^(@)`

Text Solution

Verified by Experts

The correct Answer is:
A
`/_ADC= /_ BAD= 30^(@) ` [ alt. int. `/_s]`.
`/_ ADB = 90^(@) `( angle in a semicircle).
`:./_ CDB = 30^(@) + 90^(@) = 120^(@)`
But, ABCD being a cyclic quadrilateral, we have
`/_ BAC + /_ CDB = 180^(@) implies /_ BAD + /_ CAD + /_ CDB = 180^(@)`
`implies 30^(@) + /_ CAD + 120^(@) = 180^(@) implies /_ CAD = 30^(@)`
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