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<OAQ = 90 ° because the radius is perpendicular to the tangent .
thus <OAB=90-60=30
Now OA= OB (radius)
Thus <OAB = <OBA= 30°
thus <AOB=120°
now we know that a chord subtends twice angle at the centre than it subtends at it boundary on the same side as centre.
Thus we take any point G on the boundary and connect it to A and B.
thus <AGB= 60°
But AGBC is a cyclic quadrilateral.
thus ACB = 180-60=120°.
{ u can directly use the formula also <ACB= 180- (<AOB/2)}
thus <OAB=90-60=30
Now OA= OB (radius)
Thus <OAB = <OBA= 30°
thus <AOB=120°
now we know that a chord subtends twice angle at the centre than it subtends at it boundary on the same side as centre.
Thus we take any point G on the boundary and connect it to A and B.
thus <AGB= 60°
But AGBC is a cyclic quadrilateral.
thus ACB = 180-60=120°.
{ u can directly use the formula also <ACB= 180- (<AOB/2)}
ani99ket:
Mark the answer as brainliest if u think it really helped you.
where shoulwe point g
i have assumed a point G on the larger sector . when u join these points u get a cyclic quadrilateral
since angle subtended by a chord on a sector is always same as half of that of that subtended at the centre . all points on the larger sector are equivalent .
mark it as branliest if u got the help needed
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