Question

tell the ans. of this ques with explaination​

Answer 1

use external angle of a cyclic quadrilateral

Answer 2

Join CB
AngleACB= 1/2 of 130
ACB= 65


Extend the arc BD to the point P on the circumference of the circle


Now CBPD is a cyclic quadrilateral

Since ACD is a straight line,
So ACB + BCD = 180
65+ BCD= 180
BCD = 115

DPB+ DCB=180 ( opposite angle of cyclic quadrilateral add up to 180)

DPB=180 - 115
DPB=65


DO’B= x

X=2(DPB) ..........( angle subtended by an arc at the centre is double the angle subtended by it anywhere else)
X=2*65
X=130






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