# In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.

[Hint: Draw a line parallel to ST through point R.]

**Solution:**

Given: PQ || ST, ∠PQR = 110° and ∠RST = 130°

To Find: ∠QRS

Lines which are parallel to the same line are parallel to each other.

When two parallel lines are cut by a transversal, co-interior angles formed are supplementary.

Construction: Draw a line AB parallel to ST through point R. Since AB || ST and we know that PQ || ST. Thus, AB || PQ.

Let ∠SRQ = x, ∠SRB = y and ∠QRA = z

Lines ST and AB are parallel with transversal SR intersecting them. Therefore, the co-interior angles formed are supplementary.

∠RST + ∠SRB = 180°

130° + y = 180°

y = 180° - 130° = 50°

Thus, ∠SRB = y = 50°

Similarly, lines PQ and AB are parallel with transversal QR intersecting the two lines. Therefore, the co-interior angles are supplementary.

∠PQR + ∠QRA = 180°

110° + z = 180°

z = 180° - 110° = 70°

Thus, ∠QRA = z = 70°

AB is a line, RQ and RS are rays on AB. Hence,

∠QRA + ∠QRS + ∠SRB = 180°

70° + x + 50° = 180°

120° + x = 180°

x = 180° - 120° = 60°

Thus, ∠QRS = x = 60°.

**Video Solution:**

## In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS. [Hint: Draw a line parallel to ST through point R.]

### NCERT Solutions Class 9 Maths - Chapter 6 Exercise 6.2 Question 4:

**Summary:**

In Fig. 6.31, if PQ || ST, ∠PQR = 110°, and ∠RST = 130°, then the value of ∠QRS is calculated as 60°.