Question

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

Answer 1

Answer:

given :

angle BOC = 75°

✨ we have to find angle ODC

angle BOC = angle AOD = 75 °

( vertically opposite angle )

since diagonal of a rectangle are equal and bisect each other

➡️OA = OB = OC = OD

now in Δ AOD

= angle AOD + angle OAD + angle ODA = 180°

( angle sum property )

= 75 + x + x + =180 °

= 2 x = 180 - 75

= x = 52.5°

✨ now we know that each angle of the triangle is 90 °

so we have

= angle ODA + angle ODC= 90 °

= 52.5 + angle ODC= 90 °

= angle ODC= 37.5°

here your answer mate ✔️✨

Answer 2

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the diagonals of a rectangle ABCD intersect at O. if angle BOC = to 75 degree find angle ODC?

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Given:-∠boc=75°

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We have to find =∠ODC

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∠BOC = ∠AOD ..(Vertically Opposite Angle)

⇨∠AOD=75°

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Since Diogonal Of rectangle are Equal & They Bisect Each Other.

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⇨OA=OB=OC=OD

In ΔAOD

⇨OA=OD

⇨∠OAD=∠ODD = x

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(Angle opposite to equal side of triangle are equal)

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in ΔAOD

⇨∠AOD+∠OAD+∠ODA= 180 (Linear pair)

⇨75°+x+x = 180°

⇨2x = (180-75)°

⇨2x = 105°

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⇨ x = 52.5°

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Now We Know That each angle of rectangle is 90°.

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So we Have:-

⇨∠ODA+ ODC = 90°

⇨52.5°+ ODC = 90°

⇨∠ODC = (90°-52.5°)

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⇨∠ODC = 37.5°