Question

answer this question ​

Answer 1

In parallelogram ABCD,

Given, Angle DAB = 70°

Angle DBC = 50°

So, Angle ADB = 50° (Alternate interior angle)

Similarly, Angle BCD = 70°

Now in ∆BCD,

Angle DBC + Angle BCD + Angle CDB = 180° (angle sum property of triangle)

= 50° + 70° + Angle CDB = 180°

= 120° + Angle CDB = 180°

= Angle CDB = 180° - 120°

= Angle CDB = 60°

Answer 2

in parallelogram opposite angles are equal

/DAB = /DCB = 70°

in triangle DBC

sum of all angles of triangle is 180

50+70+ /CDB = 180

/CDB = 180-120 = 60°

corresponding angles of parallelogram are equal

/DBC =/ADB =50°