Question

The diagonals of a rectangle ABCD meet at O. If angle BOC=50° then find angle ODA.​

Answer 1

Given angle BOC = 50°

and angle AOD = BOC ...( vertically opposite angles)

now, in ∆AOD , since AO = OD ...( diagonal bisector)

so, angle ODA = OAD , let say x .

now adding all angles we get

x+x + 50° = 180°

=> 2x = 180°-50° = 130°

=>x = 130°/2 = 65°

which means angle ODA = 65°

Answer 2

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

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

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