DB is a diameter of circle O. If mACB = 30°, what is the mDBA?

Solution:

Since DB is a diameter of the circle ∠DAB = 90⁰

(As we know that angle in a semicircle is 90⁰ )

Also,

∠ADB = ∠BCA = 30⁰

(Since angle in same segment of the circle are equal).

Now from △DAB

⇒ ∠DAB + ∠ADB + ∠ABD = 180⁰

⇒ 90⁰ + 30⁰ + ∠ABD = 180⁰

⇒ ∠ABD= 60⁰

DB is a diameter of circle O. If mACB = 30°, what is the mDBA?

Summary:

DB is a diameter of circle O. If ∠ACB = 30°, what is the ∠DBA= 60⁰