Question

ABCD is a rectangle if angle oab=50 find
angle aob

Answer 1

Given in rectangle ABCD,

∠BOC = 44°

∠BOC + ∠AOB = 180° [ linear Pair]

44° +∠AOB = 180°

∠AOB = 180°- 44°

∠AOB = 136°

Since diagonal of a rectangle are equal and they bisect each other.

OA = OB = OC = OD

Hence ∆AOB is an isosceles triangle.

OA= OB

∠OAB = ∠OBA [Angles opposite to equal sides of a triangle are equal]

Let ∠OAB= ∠OBA = x

∠OAB+ ∠OBA +∠AOB = 180°

x + x + 136° = 180°

2x = 180° - 136°

2 x = 44°

x= 44/2= 22°

Hence ∠OAB = 22°

Answer 2

ANSWER:-

25°

EXPLANATION:-

Angle OAB and Angle AOB are Linear pair . So,

Angle OAB + Angle AOB = 180°

50°+ Angle AOB = 180°

Angle AOB = 180° - 50°

Angle AOB = 130°

OA=OB=OC=OD

Hence ️AOB is a ISOSCELES TRIANGLE

Angle OAB = Angle OBA [Angles opposite to equal sides of a triangle are equal]

Let angle OAB = Angle OBA = X

Angle OAB + Angle OBA + Angle AOB = 180°

X+X+130°=180°

2X=180-130°

2X=50°

X=50°/2

[X=25°]