Question

the diagonals of rectangle ABCD intersect each other at O . if angle AOD is 30 then find angle OCD. pls anwer

Answer 1

Given the parallelogram is a rectangle.

In ∆OBC
angle BOC = 30° (vertically opposite)
OB = OC (diagonals are equal and bisect eachother)

So using isoceles triangle property we have
angle OBC = angle OCB

Let it be x
so by angle sum property
30° + x + x = 180°
=> 2x = 180° - 30 = 150°
x = 150/2
x = 75°


Now we know that each angle of a rectangle is 90°. So we have angle OCD + OCB = 90°
and angle OCB = 75°
so OCD + 75= 90°
OCD = 90 - 75
OCD = 15°

So we have angle OCD = 15°

Hope it helps dear friend ☺️✌️✌️

Answer 2

● Given in Rectangle ABCD

=> angle(AOD) = 30

____________________________

We know that

=> angle(AOD) + angle(DOC) = 180
____________________[ linear pair ]

=> 30 + angle(DOC) = 180 _____[given]

=> angle(DOC) = 180 - 30

=> angle(DOC) = 150 _________Eq(1)

_____________________________

{◢ Since diagonal of a Rectangle are equal and they bisect each other }

Then,

=> OA = OB = OC = OD

______________________________

Now,

△DOC

{◢ Angles opposite to equal sides of a triangle are equal }

Then,

=> angle(ODC) = angle(OCD)

Let's

=> angle(ODC) = angle(OCD) = X

______________________________

Again In △DOC

We know that,

=> angle(DOC) + angle(ODC) + angle(OCD) = 180

______________________________

Plug the all values we get,

=> 150 + X + X = 180

=> 150 + 2X = 180

=> 2X = 30

=> X = 15

_______________________________

HENCE,

=> ANGLE(OCD) = 15 __________________ANSWER

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$BE \: \: BRAINLY$