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ABCDis a rectangle whose diagonals AC and BD intersect at O. If angle OCD=62, Find angle ODA



We have, ABCD as the given rectangle in which AC and BD are the diagonals that intersect at O.We know that diagonals of rectangle bisct each other, so,OD = OB = BD2  .....1 and  OA =OC = AC2 .......2now, in rectangle diagonals are equal , soAC = BDFrom 1 and 2, we get2OD = 2OA⇒OD = OAIn ∆ODC,OD = OA⇒∠OCD = ∠ODC  Angles opposite to equal sides are equal⇒∠ODC = 62°Now, each angle of rectangle is a right angle.So, ∠D = 90°⇒∠ADB + ∠ODC = 90°⇒∠ADB +62° = 90°⇒∠ADB = 28°

 

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Given angle OCD = 62
Therefore in triangle ODC,
OD = OC (since, diagonals of a rectangle are equal to each other and they bisect each other)
 Therefore,
angle OCD = angle ODC = 62
Now angle ODC + angle ADB = 90  (sine each angle of rectangle is 90)
                 62       +     ADB      =  90
                                angle ADB  = 26
                            angle ODA = 26

 
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