Question

the diagonals of a rectangle ABCD intersect at point O. If angle COD = 78°, then find the value of angle OAB​

Answer 1

Answer:

angle AOB = angle DOC

In ∆ OAB

angle O + angle A + angle B = 180

78 + x + x = 180

78 + 2x = 180

2x = 180 - 78

2x = 102

x = 102/2

x = 51

angle OAB = 51

Step-by-step explanation:

Answer 2

Given: The diagonals of a rectangle ABCD intersect at point O and angle COD = 78°.

To find: The value of angle OAB​.

Solution:

In a rectangle, when the diagonals are intersected, the opposite angles formed at the intersection are equal. So, the angles COD and AOB are equal and the angles AOC and BOD are equal.

$\angle COD = \angle AOB$

           $= 78$

$\angle AOC = \angle BOD$

Now, in the triangle AOB formed, ∠A and ∠B are equal as ΔAOB is isosceles. The triangles formed by the intersection of the diagonals of a rectangle are always isosceles. Let ∠A and ∠B be equal to x° each. The sum of all angles of a triangle is 180°.

$\angle A + \angle B + \angle O = 180$

$x + x + 78 = 180$

$2x = 102$

$x = 51$

Therefore, the value of angle OAB​ is 51°.