Question

The diagonals of a rectangle ABCD meet at O . Angle BOC= 50.Then find angle ODA.

Answer 1

$\angle ODA=65^{\circ}$

Step-by-step explanation:

Angle BOC=50 degrees

Let angle OBC=x

Diagonals of rectangle are equal

AC=BD

Diagonals of rectangle bisect to each other.

OD=OB

OA=OC

$\frac{1}{2}AC=\frac{1}{2}BD$

OC=OB

Angle OBC=Angle OCB=x

Angle made by two equal sides are equal.

In triangle OBC

Angle OBC+angle OCB+angle BOC=180 degrees

Substitute the values then we get

x+x+50=180

2x=180-50=130

$x=\frac{130}{2}=65^{\circ}$

$\angle OBC=\angle OCB=65^{\circ}$

We know that rectangle is also a parallelogram

AB is parallel to CD

Angle OBC=Angle ODA

Reason: Alternate interior angles are equal

$\angle ODA=65^{\circ}$

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Answer 2

thank you ! ! !!!!!!!!!!