Question

can somebody plz help me....
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Answer 1

In the given figure,

Angel DBC = 80°

ED is parallel to AC

∆EDB is an isosceles triangle since both lines EB and DB are equal as they are both emerging from the ends of chord ED and connecting at the same point.

Now, since EB and AC are parallel, Angle EBA = Angle DBC

Therefore, Angle EBA + Angle EBD + Angle DBC = 180° ( Angles in a straight line equals 180°)

Substituting the values:

80° + 80° + Angle EBD = 180°

Angle EBD= 180° - ( 80°+80°)

Angle EBD = 180° - 160°

Angle EBD = 20°