# In Fig. 6.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

**Solution:**

Given: ∠SPR = 135° and ∠PQT = 110°

To find: ∠PRQ

We know that if non-common arms of two adjacent angles form a line, then these angles are called linear pair angle and their sum is equal to 180°.

If the sum of two adjacent angles is 180° then the two non-common arms of the angles form a line.

According to Angle sum property of a triangle , sum of the interior angles of a triangle is 180°.

∠SPR + ∠QPR = 180° [Linear pair]

135° + ∠QPR = 180°

∠QPR = 180° - 135°

∠QPR = 45°.....(i)

∠PQT + ∠PQR = 180° [Linear pair]

110° + ∠PQR = 180°

∠PQR = 180° - 110°

∠PQR = 70°.....(ii)

Now,

∠PQR + ∠QPR + ∠PRQ = 180° [Angle sum property of a triangle]

70°+ 45° + ∠PRQ = 180° [from (i) and (ii)]

∠PRQ = 180° - 115°

∠PRQ = 65°

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

**Video Solution:**

## In Fig. 6.39, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

NCERT Maths Solutions Class 9 Chapter 6 Exercise 6.3 Question 1

**Summary:**

In Fig. 6.39, sides QP and RQ of ΔPQR are produced to points S and T respectively, if ∠SPR = 135° and ∠PQT = 110°, then the value of ∠PRQ = 65°.

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

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