Question

the diagonal of a rectangle abcd intersect at o if angle boc=44 find angle oab

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°

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Hope this will help you...

Answer 2

Answer:

Step-by-step explanation:

∠BOC = ∠AOD� [Vertically opposite angles are equal] ∠AOD = 70° Since diagonal of a rectangle are equal and they bisect each other. We can write OA = OB = OC = OD Hence�DAOD is an isosceles triangle. ⇒∠OAD = ∠ODA� [Angles opposite to equal sides of a triangle are equal] Let ∠OAD = ∠ODA = x ∠OAD + ∠ODA +∠AOD = 180° �⇒�x + x + 70° = 180° ⇒�2x = 110° �⇒�x = 55° Hence ∠ODA = 55