Diagonals of a parallelogram ABCD intersect a | vedclass

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(A) In a parallelogram $ABCD,$ the diagonals $AC$ and $BD$ intersect at $O.$Given $\angle BOC = 90^{\circ}.$ Since $AC$ is a straight line,$\angle BOC

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) In a parallelogram $ABCD,$ the diagonals $AC$ and $BD$ intersect at $O.$Given $\angle BOC = 90^{\circ}.$ Since $AC$ is a straight line,$\angle BOC + \angle AOB = 180^{\circ},$ so $\angle AOB = 90^{\circ}.$In $	riangle BOC,$ the sum of angles is $180^{\circ}.$ Thus,$\angle OBC + \angle BOC + \angle OCB = 180^{\circ}.$Since $AB \parallel DC,$ alternate interior angles are equal,so $\angle OAB = \angle OCD.$Also,$\angle BDC = 50^{\circ}.$ Since $AB \parallel DC,$ $\angle ABD = \angle BDC = 50^{\circ}$ (alternate interior angles).In $	riangle AOB,$ $\angle AOB = 90^{\circ}$ and $\angle ABO = 50^{\circ}.$The sum of angles in $	riangle AOB$ is $180^{\circ},$$\angle OAB + \angle AOB + \angle ABO = 180^{\circ}$$\angle OAB + 90^{\circ} + 50^{\circ} = 180^{\circ}$$\angle OAB + 140^{\circ} = 180^{\circ}$$\angle OAB = 180^{\circ} - 140^{\circ} = 40^{\circ}.$

Diagonals of a parallelogram $ABCD$ intersect at $O.$ If $\angle BOC = 90^{\circ}$ and $\angle BDC = 50^{\circ},$ then $\angle OAB$ is (in $^{\circ}$)

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