Question

diagonals of a parallelogram ABCD intersect at o if angle boc = 90degree and angle BDC = 50degree find angle OAB

Answer 1

ABCD is a //gm
Since, AB//CD
angle BDC=angle ABD
Angle ABD=50°
Then,
angle BOC+angleAOB=180°
=)90°+angle AOB=180°
=)angleAOB=90°
Now,
In triangle AOB,
=)50°+90°+angle OAB=180°
=)angle OAB=180°-140°
Hence, angle OAB=40°
Hope it helps u.

Answer 2

Given:

Angle BOC=90°

Angle BDC=50°

To find:

Angle OAB

Solution:

The value of angle OAB=40°.

We can find the measure of angle by following the steps given-

We know that the diagonals of the parallelogram are intersecting at 90°.

Since the diagonals are straight lines, angle DOC is also equal to 90°.

Also, angle ODC=50°. (Given)

Now, in triangle DOC,

angle DOC+ angle ODC+ angle OCD=180° (Sum of all angles of a triangle)

On putting the values, we get

90°+50°+angle OCD=180°

140°+Angle OCD=180°

Angle OCD=180°-140°

Angle OCD=40°

Since ABCD is a parallelogram, AB is parallel to CD and AC is the transversal.

The alternate interior angles formed between parallel lines are of equal measure.

So, angle OAB=angle OCD (Alternate interior angles are equal)

Angle OAB=40°.

Therefore, the value of angle OAB=40°.