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ABCD is a cyclic trapezium with AD || BC. If ∠A = 105°, then other three angles are
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- ∠B = 75°, ∠C = 75°, ∠D = 105°
- ∠B = 105°, ∠C = 75°, ∠D = 75°
- ∠B = 75°, ∠C = 105°, ∠D = 75°
- ∠B = 105°, ∠C = 105°, ∠D = 75°
- ∠B = 75°, ∠C = 75°, ∠D = 105°
Correct Option: A
On the basis of question we draw a figure of trapezium ABCD with AD || BC , 
AD || BC
In concyclic quadrilateral ABCD,
∠A + ∠C = 180°
⇒ 105° + ∠C = 180°
⇒ ∠C = 180° – 105° = 75°
Again,
∠A + ∠B = 180°
⇒ 105° + ∠B = 180°
⇒ ∠B = 180° – 105° = 75°
∴ ∠D = 180° – 75° = 105°
