AB is a chord in the minor segment of a circle with centre O. C is a point between A and B on the minor arc AB. The tangents to the circle at A and B meet at the point D. If ∠ACB = 116°, then the measure of ∠ADB is
Answer & explanation
Correct answer: option 4
Concept Used
When a quadrilateral is inscribed in a circle, the opposite angles of it are supplementary angles.
The angle subtended at the center is always double the angle subtended at the remaining arc.
The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.
Calculations
Point P is taken on the major arc of the circle.
Then, A and P, B and P, C and B, and A and C are joined.
From cyclic quadrilateral APBC
\(\angle\)ACB = \({116}^\circ\)
Now, \(\angle\)APB = (180 - 116) = \({64}^\circ\)
Now, \(\angle\)AOB = (64 x 2) = \({128}^\circ\)
Since OA = OB = radius of the circle
So, from quadrilateral AOBD
\(\angle\)ADB = \({360}^\circ\) - (\(\angle\)OBD + \(\angle\)OAD + \(\angle\)AOB)
\(\angle\)ADB = \({360}^\circ\) -(\({90}^\circ\) + \({90}^\circ\) + \({128}^\circ\))
\(\angle\)ADB = \({52}^\circ\)
Therefore, \(\angle\)ADB is \({52}^\circ\)