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

In the given figure, `AOB` is a diameter and `ABCD` is a cyclic quadrilateral. If `/_ ADC = 120^(@)` then `/_ BAC = ?`

A

`60^(@)`

B

`30^(@)`

C

`20^(@)`

D

`45^(@)`

Text Solution

Verified by Experts

The correct Answer is:
B
`/_ ABC + /_ ADC = 180^(@)( `opp. `/_ s` of cyclic quad. )
`implies /_ ABC + 120^(@) = 180^(@) implies /_ ABC = 60^(@)`
Also, `/_ ACB = 90^(@) ( ` angle in a semicircle) .
In `Delta ACB , /_ BAC + /_ ACB + /_ ABC= 180^(@)`
`implies /_ BAC+ 90^(@) + 60^(@) = 180^(@) implies /_ BAC = 30^(@) `
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