In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Solution:
Given: ∠PRT = 40° , ∠RPT = 95° and ∠TSQ = 75°
To find: ∠SQT
We know that when two lines intersect each other at a point then there are two pairs of vertically opposite angles formed that are equal.
According to the angle sum property of a triangle, sum of the interior angles of a triangle is 360°.
In △PRT,
∠PTR + ∠PRT + ∠RPT = 180° [Angle sum property of a triangle]
∠PTR + 40° + 95° = 180°
∠PTR = 180° - 135°
∠PTR = 45°
Now,
∠QTS = ∠PTR [Vertically opposite angles]
∠QTS = 45°...... (i)
In △TSQ,
∠QTS + ∠TSQ + ∠SQT = 180° [Angle sum property of a triangle]
45° + 75° + ∠SQT = 180° [From (i)]
∠SQT = 180° - 120°
∠SQT = 60°
Hence, ∠SQT = 60°
☛ Check: NCERT Solutions for Class 9 Maths Chapter 6
Video Solution:
In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
NCERT Maths Solutions Class 9 Chapter 6 Exercise 6.3 Question 4
Summary:
In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95°, and ∠TSQ = 75°, then ∠SQT = 60°.
☛ Related Questions:
- 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.
- 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. 6.43, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65° then find the values of x and y.