Question

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

Answer 1

Step-by-step explanation:

let angle OAD=x

OAD=ADO=x.

angle AOD =BOC=50

in triangle AOD

angle OAD + angle ODA +angle AOD =180

x + x +50= 180

2x = 130

x=65

plz mark as brainliest

and happy Diwali

Answer 2

Answer:

∠ODA=65

∘

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

2

1

AC=

2

1

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}x=

2

130

=65

∘

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

∘

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}∠ODA=65

∘