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In the adjoining figure, O is the centre...

In the adjoining figure, O is the centre of the circle and AB is one of its diameter . If `angleADC = 120^(@)` , then the value of `angleBAC` is

A

`50^(@)`

B

`60^(@)`

C

`30^(@)`

D

`40^(@)`

Text Solution

Verified by Experts

ABCD is a cyclic quadrilateral `thereforeangleABC+angleADC=180^(@)C` (by theorem 38)
or , `angleABC+120^(@)C=180^(@)"or",angleABC=180^(@)C-120^(@)C=60^(@)`
Again , AOB is a diameter. `thereforeangleACB` is a semicircular angle,
`thereforeangleACB=90^(@)` [by theorem 35]
Then in `DeltaABC,angleBAC+angleABC+angleACB=180^(@)`
or, `angleBAC+60^(@)+90^(@)=180^(@)[becauseangleABC=60^(@)andangleACB=90^(@)]`
or, `angleBAC=180^(@)-150^(@)=30^(@)`
Hence (c) is correct.
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