Answer:
x = 150°
Step-by-step explanation:
From the picture attached,
Segment AC and segment BC are the tangents to circle O.
By the property of angle between tangent and the radius of a circle,
"Angle between the radius and tangent drawn to the circle is 90°"
m(∠OAC) = m(∠OBC) = 90°
Sum of opposite angles of the quadrilateral OBCA is 180°,
m(∠OAC) + m(∠OBC) = 180°
Therefore, quadrilateral OBCA is a cyclic quadrilateral.
By the same property of cyclic quadrilateral,
m(∠AOB) + m(∠ACB) = 180°
x° + 30° = 180°
x = 180 - 30
x = 150°
