Question

The diagonals of a rectangle ABCD intersect at o. if, angle BOC = 70 degree find angle ODA

Answer 1

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°

 

Answer 2

Refer to the figure above.
$\angle BOC= \angle AOD\\ In \: \Delta AOD, \\ \angle AOD = \pi -2\angle OAD\\ 2\angle OAD=180- 70\\ \angle OAD= \frac{110}{2}\\ \angle OAD=55$