DB is a diameter of circle O. If mACB = 30°, what is the mDBA?
30°
50°
60°
70°
90°
![DB is a diameter of circle O If mACB 30 what is the mDBA 30 50 60 70 90 class=](https://us-static.z-dn.net/files/dc0/aabc2ca5671a10243298b1fcf85db039.jpg)
Answer:
m∠ABD = 60°
Step-by-step explanation:
From the given figure :
Given : m∠ACB = 30°
To find : m∠DBA
Solution : Since, the angle formed in the same segment are equal in the circle
⇒ m∠ACB = m∠ADB
Since, m∠ACB = 30°
⇒ m∠ADB = 30°
Now, Angle formed in the semicircle is right angle
⇒ m∠BAD = 90°
So, by using angle sum property of a triangle in the ΔADB. We get,
m∠ADB + m∠ABD + m∠BAD = 180
⇒ 30 + m∠ABD + 90 = 180
⇒ m∠ABD + 120 = 180
⇒ m∠ABD = 60°