Given : $\angle \mathrm{ABC}={100}^{\circ },\angle \mathrm{ACD}={40}^{\circ }$
We know that,
$\angle \mathrm{ABC}+\angle \mathrm{ADC}={180}^{\circ }$
(The sum of opposite angles in a cyclic quadrilateral $={180}^{\circ }$ )
$\therefore {100}^{\circ }+\angle \mathrm{ADC}={180}^{\circ }$
$\angle \mathrm{ADC}={180}^{\circ }-{100}^{\circ }$
$\angle \mathrm{ADC}={80}^{\circ }$
Join OA and OC, we have a isosceles $△\mathrm{OAC}$ ,

$\because \mathrm{OA}=\mathrm{OC}$
$\therefore \angle \mathrm{AOC}=2\times \angle \mathrm{ADC}$ (by theorem)
$\angle \mathrm{AOC}=2\times {80}^{\circ }={160}^{\circ }$
In $△\mathrm{AOC}$ ,
$\angle \mathrm{AOC}+\angle \mathrm{OAC}+\angle \mathrm{OCA}={180}^{\circ }$
${160}^{\circ }+\angle \mathrm{OCA}+\angle \mathrm{OCA}={180}^{\circ }$
$\left[\because \angle \mathrm{OAC}=\angle \mathrm{OCA}\right]$
$2\angle \mathrm{OCA}={20}^{\circ }$
$\angle \mathrm{OCA}={10}^{\circ }$
$\angle \mathrm{OCA}+\angle \mathrm{OCD}={40}^{\circ }$
${10}^{\circ }+\angle \mathrm{OCD}={40}^{\circ }$
$\therefore \angle \mathrm{OCD}={30}^{\circ }$
$\mathrm{Hence},\angle \mathrm{OCD}+\angle \mathrm{DCT}=\angle \mathrm{OCT}$
$\because \angle \mathrm{OCT}={90}^{\circ }$
(The tangent at a point to circle is $\mathit{⟂}$ to the radius through the point of contact)
${30}^{\circ }+\angle \mathrm{DCT}={90}^{\circ }$
$\therefore \angle \mathrm{DCT}={60}^{\circ }$