$BA$ and $BC$ are tangents to the given circle (with center $O$) from the point $B$.
$\angle AOB = 65^\circ$. Find $\angle AOC$, $\angle ADC$ and $\angle ABC$.
If two tangents $BA$ and $BC$ are drawn to a circle (with center $O$) from an exterior point ($B$), then
- the two tangents are equal in length $\;$ i.e. $\;$ $BA = BC$
- the tangents subtend equal angles at the center of the circle $\;$ i.e. $\;$ $\angle AOB = \angle COB$
- the tangents are equally inclined to the line joining the point $\left(B\right)$ and the center of the circle $\left(O\right)$, $\;$ i.e. $\;$ $\angle ABO = \angle CBO$
$\angle AOC = \angle AOB + \angle COB = 65^\circ + 65^\circ = 130^\circ$
Now, $\angle AOC = 2 \angle ADC$ $\;\;$ [angle at the center of a circle is twice the angle at remaining circumference]
$\therefore \;$ $\angle ADC = \dfrac{\angle AOC}{2} = \dfrac{130^\circ}{2} = 65^\circ$
$OA = OC = $ radii of the same circle
$OA \perp AB$; $\;$ $OC \perp CB$ $\;\;\;$ [the tangent at any point of a circle and the radius through this point are perpendicular to each other]
$\therefore \;$ $\angle OAB = \angle OCB = 90^\circ$
In right $\triangle OAB$, $\;$ $\angle OAB + \angle AOB + \angle ABO = 180^\circ$ $\;\;$ [sum of angles of a triangle]
i.e. $\;$ $90^\circ + 65^\circ + \angle ABO = 180^\circ$
$\implies$ $\angle ABO = 25^\circ$
Now, $\angle CBO = \angle ABO = 25^\circ$ $\;\;$ [by (3.)]
$\therefore \;$ $\angle ABC = \angle ABO + \angle CBO = 25^\circ + 25^\circ = 50^\circ$
$\therefore \;$ $\angle AOC = 130^\circ$, $\;$ $\angle ADC = 65^\circ$, $\;$ $\angle ABC = 50^\circ$