the diagonals of rectangle abcd intersect at o . if boc= 70 degree then find oda and abo
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Step-by-step explanation:
Given in rectangle ABCD, ∠BOC = 70°
∠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°
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Answered by
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Given in rectangle ABCD, ∠BOC = 70°
∠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°
∠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°
thnku!!
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