Question

In a rectangle ABCD..diagonals intersect at O.if angleAOB=118 find angleABO, angleADO and angleOCB​

Answer 1

Answer:

angleABO = 45°, angleADO= 45° and angleOCB= 45° by Dividing 2 because angle of rectangle is 90°

Answer 2

Answer: ∠ABO=31° , ∠ADO=∠OCB = 59°

Step-by-step explanation:

In a rectangle, the diagonals bisect each other

ie, AO=OB

OD=OC

In ΔAOB,

given ∠AOB = 118

∵ AO=OB,

∠BAO= ∠ABO (anlges oppostie to equal sides are equal)

also, ∠BAO + ∠ABO + 118 = 180

or 2∠ABO = 180-118

∠ABO = 31°

this means that ∠ODC = 31° (because AB║DC and diagonal BD is the transversal, ∴ ∠ABO=∠ODC)

∴∠ADO = 90-∠ODC = 90-31=59

finally ∠ADO=∠OCB because ΔAOD ≅ ΔBOC (both triangles are congruent ∵ OD = OB(diagonals bisect each other) , BC=AD (opposite sides of rectangle) , AO = OC (diagonals bisect each other) )

∴ ∠OCB = 59°