In the figure, $O$ is the center of a circle.
$CT$ is a tangent to the circle at $C$.
Find $\angle ADC$ and $\angle DCT$.
$O$ is the center of the circle; $\;$ $\angle ABC = 100^\circ$; $\;$ $\angle ACD = 40^\circ$;
$CT$ is a tangent to the circle at a point $C$
In the figure, $\;$ $ABCD$ is a cyclic quadrilateral.
$\therefore \;$ $\angle ABC + \angle ADC = 180^\circ$ $\;\;\;$ [opposite angles of a cyclic quadrilateral are supplementary]
$\implies$ $\angle ADC = 180^\circ - \angle ABC = 180^\circ - 100^\circ = 80^\circ$
Now, $\;$ $\angle AOC = 2 \angle ADC$
[angle at the center of a circle is twice the angle at the remaining circumference]
$\therefore \;$ $\angle AOC = 2 \times 80^\circ = 160^\circ$
Join $OA$ and $OC$.
$\implies$ $\angle ACO = \angle CAO$ $\;\;\;$ [angles opposite equal sides are equal]
In $\triangle AOC$, $\;$ $\angle AOC + \angle ACO + \angle CAO = 180^\circ$ $\;\;\;$ [sum of angles of a triangle]
i.e. $\;$ $\angle AOC + 2 \angle ACO = 180^\circ$
$\implies$ $\angle ACO = \dfrac{180^\circ - \angle AOC}{2} = \dfrac{180^\circ - 160^\circ}{2} = 10^\circ$
From the figure, $\;$ $\angle ACD = \angle ACO + \angle OCD$
$\therefore \;$ $\angle OCD = \angle ACD - \angle ACO = 40^\circ - 10^\circ = 30^\circ$
Also, $\;$ $\angle OCT = 90^\circ$
[tangent at any point of a circle and the radius through this point are perpendicular to each other]
But, $\;$ $\angle OCT = \angle OCD + \angle DCT$
$\therefore \;$ $\angle DCT = \angle OCT - \angle OCD = 90^\circ - 30^\circ = 60^\circ$