Question

pls solve this for me ....... pls!!!!!



in a parallelogram ABCD, angle B is 50° more than angle A. what is the measure of the angles of ABCD????

Answer 1

We know that sum of angles of a parallelogram is 180.

= > A + B= 180    ---- (1)

Given that Angle B is 50 more than Angle A.

= > B = A + 50   ----- (2)

Substitute (2) in ((1), we get

= > A + B = 180

= > A + A + 50 = 180

= > 2A + 50 = 180

= > 2A = 180 - 50

= > 2A = 130

= > A = 65.

Substitute A = 65 in (2), we get

= > B = A + 50

= > B = 65 + 50

= > B = 115.

We know that in a parallelogram opposite sides are equal, So

A = C = 65

B = D = 115.

Hope this helps!

Answer 2

let angle A=x
angle B=50+x

angle A= angle C (opp. angle of parallelogram
are equal)
angle B=angle D

A+B+C+D=360 ( sum of all angles of
quadrilateral)

x+ 50+x +x +50+x=360
4x +100=360
4x= 360-100
4x = 260
x = 260/4
x = 65

hence, angle A= 65°
B = 65+50 =115°
angle A= angle C= 65°
angle B= angle D = 115°