In the given figure, is a diameter of a circle with center If and are straight lines, meeting at such that and find

(i) (ii) (iii)

(i)

∠BDA = 90°= ∠EDB [Semi circle angle]

In triangle EBD,

∠DBE + ∠EDB + ∠BED = 180°

⇒ ∠DBE + 90° + 25° = 180°

⇒ ∠DBE + 115° = 180°

⇒ ∠DBE = 65°

Now,

∠DBC + ∠DBE = 180°[CBE is a straight line]

⇒ ∠DBC + 65° = 180°

⇒ ∠DBC = 115°

(ii)

∠DCB = ∠BAD [Angle in the same segment]

∴∠DCB = 35°

(iii) ∠BDC = 30°

In triangle BCD,

∠BDC + ∠DCB + ∠DBC = 180°

⇒ ∠BDC + 35° + 115° = 180°

⇒ ∠BDC + 150° = 180°

⇒ ∠BDC = 30°