Question

In the figure XY and X¹Y¹ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X¹Y¹ at B then <AOB =?? please answer this question please friends just find only <AOB no need to prove..... ​

Answer 1

Answer:

Step-by-step explanation:

OP  xy (tangent  radius)

OC  AB

In OPA and OCA

OPA = OCA = 90o

OP = OC (radii)

AP = AC (tangents from an external point)

OPA  OCA (SAS)

1 = 2 (CPCT)

2 = PAC

Similarly, 3 = 4

3 =  QBC

xy ||x'y' and AB is transversal

PAB + QBA = 180o (interior angles on same side of transversal)

Or PAC = QBC = 180o

2 + 3 = 90o

In OAB,

AOB + 2 + 3 = 180o

AOB = 90o

Hence proved.