Given-Angle QPO = angle PRQ( alternate interior angles)

Also, QOR =90( the diagonal bisect each other at 90 degree)

In triangle QRO,

The sum of all angles of a triangle= 180

SQR+PRQ+QOR=180

SQR+ 30+90=180

SQR+120= 180

SQR=180-120

SQR =60

I hope this helps you

Verified answer

Step-by-step explanation:

We know that opposite angles in a parallelogram are equal. Therefore, angle QPS is also 30°.

Since the diagonals PR and QS of the parallelogram intersect at O, we have:

angle QPO + angle SPO = 180° (sum of angles in triangle POS)

angle QPS + angle SPO = 180° (sum of angles in triangle PQO)

Substituting the values we know, we get:

30° + angle SPO = 180° (1)

30° + angle SPO = 180° (2)

From equations (1) and (2), we get:

angle SPO = 150°

Now, considering the triangle SQR, we know:

angle SQR + angle QRS + angle RSQ = 180° (sum of angles in triangle SQR)

Substituting the values we know, we get:

angle SQR + 30° + 30° = 180°

Simplifying, we get:

angle SQR = 120°

Therefore, the answer is (d) 100°.

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