SSC CHSL Model Paper 15

ABCD Is a concyclic quadrilateral.
∠ACB = Angle in semi-circle = 90°
∴ ∠ABC = 90° - ∠CAB = 90° - 25° = 65°
∴ ∠ABC + ∠ADC = 180°
⇒ ∠ADC = 180° - 65° = 115°
DC I I AB
∴ ∠ACD = ∠CAB = 25°
In ΔACD,
∠CAD + ∠ADC + ∠ACD = 180°
⇒ ∠CAD + 115° + 25° = 180°
⇒ ∠CAD = 180° - 140° = 40°