Question

In a rectangle ABCD, AC and BD intersect at
O. angle BOC = 40°. Find angle OBC and angle OCB.​

Answer 1

Answer :-

• ∠BOC = 40°

• ∠OBC = 70°

• ∠OCB = 70°

Step-by-step explanation :-

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Solution :-

in rectangle ABCD,

AC = BD ( diagonals of rectangle are equal )

------------------------

in ΔBOC,

∠BOC = 40°  ( given ) -----1

∠OBC = ∠OCB ( base angle theorem ) -----2

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∠OBC  + ∠OCB  +  ∠BOC  = 180° ( angle sum property )

∠OBC  + ∠OBC  + 40° = 180° ( from 2 )

2∠OBC = 180 - 40

2∠OBC =  140

∠OBC = 140/2

∠OBC = 70

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Therefore,

∠OBC = 70 and ∠OCB = 70°

Answer 2

Answers:

∠OBC = 70°

∠OCB = 70°

∠OBC = ∠OCB

Step-by-step explanation:

Given Problem:

In a rectangle ABCD, AC and BD intersect at

O. angle BOC = 40°. Find angle OBC and angle OCB.​

Solution:

To Find:

Angle OBC and angle OCB.​

-------------------

Method:

Given that,

In rectangle ABCD,  

Diagonals of rectangle are equal:

AC = BD

Now,

In ΔBOC,  

∠BOC = 40°..................Eq(1)

We know that,

Base angle theorem.

So,

∠OBC = ∠OCB..........Eq(2)

Now,

We know that,

Angle sum property.

So,

∠OBC  + ∠OCB  +  ∠BOC  = 180°    (Here is Angle sum property)

∠OBC  + ∠OBC  + 40° = 180°          (From Eq2)

2∠OBC = 180 - 40

2∠OBC =  140

∠OBC = $\dfrac{140}{2}$

∠OBC = 70

Hence,

∠OBC = 70 and ∠OCB = 70°