Given : $\angle ABC=100^{\circ}, \angle ACD=40^{\circ}$We know that, $\angle ABC+\angle ADC=180^{\circ}$(The sum of opposite angles in a cyclic quadrilateral $=180^{\circ}$ ) $\therefore \;\; 100^{\circ}+\angle ADC\;=180^{\circ}$ $\angle ADC\;=180^{\circ}-100^{\circ}$ $\angle ADC\;=80^{\circ}$Join OA and OC, we have a isosceles $\triangle OAC$ , $\; \because \;\; OA=OC$ $\;\;\text{(Radii of a circle)}\;$ $\; \therefore \;\; \angle AOC=2 \times \angle ADC$ (by theorem) $\angle AOC=2 \times 80^{\circ}=160^{\circ}$In $\triangle AOC$ , $\angle AOC+\angle OAC+\angle OCA\;=180^{\circ}$ $160^{\circ}+\angle OCA+\angle OCA\;=180^{\circ}$ ${[ \because \angle OAC=\angle OCA]}\;$ $2 \angle OCA\;=20^{\circ}$ $\angle OCA\;=10^{\circ}$ $\angle OCA+\angle OCD\;=40^{\circ}$ $10^{\circ}+\angle OCD\;=40^{\circ}$ $\therefore\angle OCD\;=30^{\circ}$ $Hence,\;\; \angle OCD+\angle DCT\;=\angle OCT$ $\because \angle OCT\;=90^{\circ}$(The tangent at a point to circle is $\perp$ to the radius through the point of contact) $\;30^{\circ}+\angle DCT\;=90^{\circ}$ $\; \therefore \;\; \angle DCT\;=60^{\circ}$