Given: BD is a diameter
m 1 = 100°
m BC= 30°
m ADB=
160
280
330

Answer with explanation:
Given
m∠1=100°
[tex]m \widehat{BC}=30 ^{\circ}[/tex]
To Find:---m∠A D B
Solution
In Δ A OD
Represent the center of circle by O.
→ m∠1=100°
→OD=O A----------Radii of Circle
→∠ADO=∠D A O--------If opposite sides are equal angle opposite to them are equal.
In ΔA OD, Using Angle Sum property of Triangle
→∠ADO+∠D A O+∠A OD=180°
→2 ∠ADO+100° =180°-------------------[∠ADO=∠D A O]
→2∠ADO=180° -100°
→2∠ADO=80°
Dividing both sides by , 2 we get
⇒∠ADO=40°
⇒⇒⇒∠A DB=40°
≡⇒If you are asking about
[tex]m \widehat{ADB}=180 ^{\circ}[/tex]
Because Angle in a semicircle is Right Angle.Diameter B D divides the circle into two equal arc measure of each arc being 180 degree.
⇒⇒If you are asking about Angle made by Major arc ADB, then
[tex]m \widehat{ADB}=\angle AOD + \widehat{BOD}\\\\=100 ^{\circ}+180 ^{\circ}=280 ^{\circ}[/tex]
Option B