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