Question

$\bf \underline \red {MY \: QUESTION:- }$


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

Answer 1

Answer:

It is given that Ao and BO are angle bisectorr of angle A and B

So, angleA=2 angle OAB

and angle B=2angleOBA

Sum of all angles of a quadrilateral is 360°

so

angles (A+B+C+D)=360°

(A+B+70°+50°)=360°

A+B=360°-120°

A+B=240

2angle OAB+2angleOBA=240°

2angle (OAB+OBA)= 240°

angle (OAB+OBA)=240°/2=120°

Sum of all angles of a triangle is 180°

so, angle (OAB+OBA+AOB)=180 °

120°+AOB=180°

angleAOB=(180-120)°=60°

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Answer 2

Step-by-step explanation:

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°

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