Question

Q.3 Diagonals AC and BD of parallelogram ABCD intersect at O. If _ <BOC = 90° and <BDC = 50°, find <OAB. ​

Answer 1

Answer:

Given:-

Angle BOC= 90°

Angle BDC= 50°

Step-by-step explanation:

Angle BOC+ angle BOA= 180° (Liner pair)

90°+Angle BOA= 180°

Angle BOA= 180°-90°= 90°

Angle BDC= Angle DBA= 50°(Opposite angle equal)

Now In triangle OAB,

Angle OAB + Angle AOB + Angle OBA= 180° (Angle sum property of triangle)

Angle OAB+ 90°+50°= 180°

Angle OAB = 180°-140°

Angle OAB= 40°

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Answer 2

Step-by-step explanation:

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