Given : ∠ABC=100∘,∠ACD=40∘
We know that,
∠ABC+∠ADC=180∘
(The sum of opposite angles in a cyclic quadrilateral =180∘ )
∴‌‌100∘+∠ADC‌=180∘
∠ADC‌=180∘−100∘
∠ADC‌=80∘
Join OA and OC, we have a isosceles △OAC ,

‌∵‌‌OA=OC ‌‌ (Radii of a circle) ‌
‌∴‌‌∠AOC=2×∠ADC (by theorem)
∠AOC=2×80∘=160∘
In △AOC ,
∠AOC+∠OAC+∠OCA‌=180∘
160∘+∠OCA+∠OCA‌=180∘
[∵∠OAC=∠OCA]‌
2∠OCA‌=20∘
∠OCA‌=10∘
∠OCA+∠OCD‌=40∘
10∘+∠OCD‌=40∘
∴∠OCD‌=30∘
Hence,‌‌∠OCD+∠DCT‌=∠OCT
∵∠OCT‌=90∘
(The tangent at a point to circle is ⟂ to the radius through the point of contact)
‌30∘+∠DCT‌=90∘
‌∴‌‌∠DCT‌=60∘