Since, the tangent at any point of a crcle is perpenedicular the radius throgh the point of contact.∴ ∠OAT = 90°and ∠OBT = 90°If two tangents are drawn to a circle from an external point, then they are equally inclined to the segment joining the centre to the point∴ ∠ATO = ∠BTOBut ∠ATO = 40°Now, ∠BTO = 40°In quadrilateral, ∠OBT, we have∠OAT + ∠OBT + ∠ATB + ∠AOB= 360°⇒ 90 + 90 + 80 + ∠AOB = 360°⇒ 260 + ∠AOB = 360°⇒ ∠AOB = 360° – 260° = 100°