Question

7. In the adjoining figure, the diagonals AC and BD of parallelogram ABCD intersect at O. If angle OBC = 36°, angle ODC =28º and angle AOD = 64°; find the measure of (i) angle OAD (ii) angle OCD.​

Answer 1

Given :-

ABCD is a parallelogram

Diagonals AC and BD intersect at O

∠OBC = 36°

∠ODC = 28°

∠AOD = 64°

Required to find :-

• Measure of ∠OAD ?

• Measure of ∠OCD ?

Concept used :-

Properties of a parallelogram ;

• Opposite sides are equal and parallel .

• Opposite angles are equal .

• Diagonals bisect each other .

• Sum of two adjacent angles is supplementary .

Solution :-

Given that :-

ABCD is a parallelogram

Diagonals AC and BD intersect at O

∠OBC = 36°

∠ODC = 28°

∠AOD = 64°

So,

Let consider Parallelogram ABCD

In which ;

∠CDB = ∠DBA

[ Reason : Alternate interior angles ]

So,

∠CDB = ∠DBA = 28°

Similarly ,

∠ADB = ∠DBC = 36°

Hence,

we know that ,

∠D = ∠ADB + ∠CDB

From the above

∠CDB = 28° & ∠ADB = 36°

So,

∠D = 36° + 28°

∠D = 64°

So,

∠D = 64°

However,

We know that ;

In a parallelogram the sum of two adjacent angles is supplementary

So,

∠A + ∠D = 180°

∠A + 64° = 180°

∠A = 180° - 64°

∠A = 116°

Hence,

∠A = 116°

Now consider ,

∆ DAO & ∆ BCO

In ∆DAO & ∆ BCO

∠DOA = ∠COB

[ Reason : Vertically opposite angles ]

So,

∠DOA = ∠COB = 64°

Now consider only ∆ BCO ;

In ∆ BCO

∠COB + ∠OBC + ∠OCB = 180°

[ Reason :- Angle sum property ]

64° + 36° + ∠OCB = 180°

100° + ∠OCB = 180°

∠OCB = 180° - 100°

∠OCB = 80°

Consider the parallelogram ABCD

In parallelogram ABCD ;

∠OAD = ∠OCB

[ Reason :- Alternate interior angles ]

So,

∠OAD = ∠OCB = 80°

Hence,

∠OAD = 80°

But ,

∠A = ∠DAC + ∠BAC

So,

∠BAC = ∠A - ∠DAC

∠BAC = 116° - 80°

∠BAC = 36°

So,

∠BAC = 36°

However,

∠BAC = ∠OCD

[ Reason :- Alternate interior angles ]

∠BAC = ∠OCD = 36°

∠OCD = 36°

Therefore ,

• ∠OAD = 80°

• ∠OCD = 36°