∠ABC+∠ADC=180° (opposite angles of cyclic quadrilateral) ∠ADC=180°−115°=65° ∠AOC=2∠CDA [0 being circumcentre of ∆ADC] ∠AOC=2×65=130° Now, OA = OC [radii of the given circle] So, ∠OAC=∠OCA In ∆OAC;∠OAC+∠OCA+∠AOC=180° 2∠OAC=180°−130° ∠OAC=25° Also, OA is ⊥τ to TA So, ∠CAT=∠ZOAT−∠OAC =90°−25°=65°