Question

ABCD is a quadrilateral . AO and BO are the angle bisector of angle A and B which meet at O . If angle C =70°,angle D =50° ,find angle AOB

Answer 1

In quadrilateral ABCD,

angle A + angle B + angle C + angle D = 360° (sum of all inner angles of a quadrilateral is 360°)
angle A = x + x (since AO bisects angle A)
angle B = y + y (since BO bisects angle B)

Therefore,
2x + 2y + 70° + 50° = 360°
2(x+y) + 130° = 360°
2(x+y) = 360° - 130° = 230°
x + y = 230/2 = 115° ---------- (i)

In triangle AOB,
x+y+angle AOB = 180° (Sum of all angles of a triangle is 180°)
115° + angle AOB = 180° ( eq.{i})
Therefore angle AOB = 180° - 115° = 65°

Answer 2

Answer:

60

Step-by-step explanation:

Angle A + Angle B + Angle C + Angle D= 360 degrees

Let Angle A and Angle B be 'x'

So, x+x+70+50=360

2x+120=360

2x=360-120

2x=240

x=120

Therefore, Angle A is 120

Angle B is 120

Angle C is 70

Angle D is 50

Angle AO and Angle BO bisects Angle Aa and Angle B

So, Angle A = 120\2 = 60

Angle B = 120\2 = 60

In a triangle,

Angle AOB + Angle OAB + Angle OBA = 180 degrees

Angle AOB + 60 + 60 = 180

Angle AOB + 120 = 180

Angle AOB = 180-120

Angle AOB = 60

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