# In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65° then find the values of x and y

**Solution:**

Given: PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°

To find: x and y

We know when two parallel lines are cut by a transversal, alternate interior angles formed are equal.

According to the angle sum property of a triangle, sum of the interior angles of a triangle is 360°.

Since, PQ || SR and QR is the transversal,

∠PQR = ∠QRT [Alternate interior angles]

∠PQS + ∠SQR = ∠QRT [From figure]

x + 28° = 65°

x = 65° - 28°

x = 37°

Now, in △PQS

∠PQS + ∠PSQ + ∠QPS = 180° [Angle sum property of a triangle]

37° + y + 90° = 180° [Since. ∠QPS = 90°]

y = 180° - 127°

y = 53°

Hence, x = 37° and y = 53°

**Video Solution:**

## In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65° then find the values of x and y

### NCERT Maths Solutions Class 9 - Chapter 6 Exercise 6.3 Question 5:

**Summary:**

In Fig. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28°, and ∠QRT = 65°, then the values of x and y from the given figure are 37° and 53° respectively