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1.

In the adjoining figure, O is the centre of the circle and AB is a diameter. ABCD IS A CYCLIC QUADRILATERAL . If `angleABC=65^(@),angleDAC=40^(@)` , then the value of `angleBCD` isA. `75^(@)`B. `105^(@)`C. `115^(@)`D. `80^(@)`

Answer» AB is a diameter of the circle with centre at O.
`thereforeangleACB` is a semicircular angle . `thereforeangleACB=90^(@)`
Now , ABCD is a cyclic quadrilateral , `thereforeangleADC+angleABC=180^(@)`
or,` angleADC+65^(@)=180^(@) "or",angleADC=180^(@)-65^(@)"or",angleADC=115^(@)`
`therefore"in" DeltaACD,angleACD+angleACD=angleCAD=180^(@)`
or `angleACD+115^(@)+40^(@)=180^(@)"or",angleACD=180^(@)-155^(@)"or",angleACD=25^(@)`
`thereforeangleBCD=angleACB+angleACD=90^(@)+25^(@)=115^(@)`
(c) `because` is correct.
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