Question

the diagonal of a rectangle ABCD intersect at O if angle BOC = 75 degree .find angle ODC​

Answer 1

Answer:  75/2 degree

Step-by-step explanation:

Sorry i am not able to show you diagram, but if you will draw the diagram, you will see that

angle BOC + angle DOC  =  180 degree    

(  Sum of angles lying on a  straight line is 180 degree.)

Therefore, angle DOC = 180 - 75

                   angle DOC = 105 degree   -------------(1)

In triangle DOC, angle ODC = angle OCD  ---------(2)

 (  As OD =OC, because diagonals of rectangle are equal and bisect each other.)

Therefore,  angle ODC + angle OCD + angle DOC = 180 degree

( As sum of angles in triangle is 180 degree.)

So,  by using (1) and (2),

angle ODC + angle ODC + 105 degree  =  180 degree

2 (angle ODC) = 75 degree

angle ODC = 75/2 degree   (ANSWER)

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Answer 2

Question :

The diagonal of a rectangle ABCD intersect at O if angle BOC = 75 degree .find angle ODC.

Solution :

Given : ∠BOC = 75°

We have to find ∠ODC .

∠BOC = ∠AOD ..( vertically opposite angles )

⇒∠AOD = 75°

Since diagonal of a Rectangle are equal and they bisect each other.

⇒OA = OB = OC = OD

In ΔAOD

OA =OD

⇒∠OAD=∠ODA= x

[Angles opposite to equal sides of a triangle are equal]

Now in ᐃAOD

∠AOD+∠OAD+∠ODA = 180° (linear pair)

⇒75° +x+x = 180°

⇒2x= (180-75)°

⇒2x= 105°

⇒x = 52.5°

Now we know that each angle of a rectangle is 90°.

So we have

∠ODA + ∠ODC= 90°

52.5° + ∠ODC =90°

⇒∠ODC =( 90-52.5)°

⇒∠ODC =37.5°