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# 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°

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 6

**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

**☛ Related Questions:**

- In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∠XYZ, find ∠OZY and ∠YOZ.
- In Fig. 6.41, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
- In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
- In Fig. 6.44, the side QR of ∠PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2 ∠QPR.

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