Given :-
∠C = 90°
CD is the altitude to AB.
∠A = 65° .
Solution :-
in Right angle ∆ABC , we have ,
→ ∠ACB = 90°
→ ∠CAB = 65° .
So,
→ ∠ACB + ∠CAB + ∠CBA = 180° (By angle sum Property.)
→ 90° + 65° + ∠CBA = 180°
→ 155° + ∠CBA = 180°
→ ∠CBA = 180° - 155°
→ ∠CBA = 25° .
Now, in ∆CDB ,
→ CD is the altitude to AB.
So,
→ ∠CDB = 90°
→ ∠CBD = ∠CBA = 25° .
So,
→ ∠CDB + ∠CBD + ∠DCB = 180° (Angle sum Property.)
→ 90° + 25° + ∠DCB = 180°
→ 115° + ∠DCB = 180°
→ ∠DCB = 180° - 115°
→ ∠DCB = 65° .
Now, in ∆ADC ,
→ CD is the altitude to AB.
So,
→ ∠ADC = 90°
→ ∠CAD = ∠CAB = 65° .
So,
→ ∠ADC + ∠CAD + ∠DCA = 180° (Angle sum Property.)
→ 90° + 65° + ∠DCA = 180°
→ 155° + ∠DCA = 180°
→ ∠DCA = 180° - 155°
→ ∠DCA = 25° .
Hence, from all above (Also image.) we get,
∠DBC = 25°
∠DCB = 65°
∠CDB = 90°
∠ACD = 25°
∠ADC = 90° .
