Lines and Angles (Exercise 6.3)

Lines and Angles

Chapter 6

Exercise 6.3

EX 6.3 QUESTION 1.

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

Solution:
∠TQP + ∠PQR = 180°        [Linear pair]
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110° = 70°

∠SPR + ∠QPR = 180°        [Linear pair]

⇒135° + ∠QPR = 180°

⇒ ∠QPR = 180° – 135° = 45°

In ∆PQR is produced to S.

⇒ ∠PQR + ∠QPR +∠R = 180°
⇒ 70° + 45° +∠R = 180°

⇒ 115° +∠R = 180°

⇒ ∠R =180° – 115° = 65°

EX 6.3 QUESTION 2.

In Fig. 6.40, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of Δ XYZ, find OZY and YOZ.

Solution:

Given : ∠X = 62° and ∠XYZ = 54°

In ∆XYZ, ∠XYZ + ∠YZX + ∠ZXY = 180°         [Angle sum property of a triangle]
∴ 54° + ∠YZX + 62° = 180°
⇒ ∠YZX = 180° – 54° – 62° = 64°
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
∴ ∠OYZ =½(∠YZX)=½(54°) = 27°
and ∠OZY =½(∠YZX) = ½(64°) = 32°
In ∆OYZ, ∠YOZ + ∠OYZ + ∠OZY = 180°           [Angle sum property of a triangle]
⇒ ∠YOZ + 27° + 32° = 180°
⇒ ∠YOZ = 180° -27° – 32° = 121°
Thus, ∠OZY = 32° and ∠YOZ = 121°

EX 6.3 QUESTION 3.

 In Fig. 6.41, if AB DE, BAC = 35° and CDE = 53°, find DCE.

Solution:

Given : AB || DE

∠CED + ∠BAC        [Alternate interior angles]

∠CED = 35° [Given]
In ∆CDE,  ∠CDE + ∠CED + ∠DCE = 180°           [Angle sum property of a triangle]
∴ 53° + 35° + ∠DCE =180°
⇒ ∠DCE = 180° – 88° = 92°

EX 6.3 QUESTION 4.

 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 : ∠P = 95°, ∠R = 40°

In ∆PRT,  ∠P + ∠R + ∠PTR = 180°        [Angle sum property of a triangle]
⇒ 95° + 40° + ∠PTR =180°
⇒ ∠PTR = 180° – 95° – 40° = 45°
But PQ and RS intersect at T.
∴ ∠PTR = ∠QTS              [Vertically opposite angles]
∴ ∠QTS = 45°               [ ∵ ∠PTR = 45°]
In ∆ TQS,  ∠TSQ + ∠STQ + ∠SQT = 180°          [Angle sum property of a triangle]
∴ 75° + 45° + ∠SQT = 180°                [ ∵ ∠TSQ = 75° and ∠STQ = 45°]
⇒ ∠SQT = 180° – 120° = 60°

EX 6.3 QUESTION 5.

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

Solution:

∆ QRS, the side SR is produced to T.
∴ ∠QRT = ∠RQS + ∠RSQ             [Exterior angle property of a triangle]
But ∠RQS = 28° and ∠QRT = 65°
So, 28° + ∠RSQ = 65°
⇒ ∠RSQ = 65° – 28° = 37°
Since, PQ || SR and QS is a transversal.
∴ ∠PQS = ∠RSQ = 37°               [Alternate interior angles]
⇒ x = 37°
Again, PQ ⊥ PS ⇒ AP = 90°
In ∆PQS,  ∠P + ∠PQS + ∠PSQ = 180°         [Angle sum property of a triangle]
⇒ 90° + 37° + y = 180°
⇒ y = 180° – 90° – 37° = 53°
x = 37° and y = 53°

EX 6.3 QUESTION 6.

In Fig. 6.44, the side QR of ΔPQR is produced to a point S. If the bisectors of PQR and PRS meet at point T, then prove that QTR = ½ QPR.

Solution:

In ∆PQR, side QR is produced to S, so by exterior angle property,
∠PRS = ∠P + ∠PQR
⇒ ½∠PRS = ½∠P + ½∠PQR
⇒ ∠TRS = ½∠P + ∠TQR …(1)           [∵ QT and RT are bisectors of ∠PQR and ∠PRS respectively.]
In ∆QRT,  ∠TRS = ∠TQR + ∠T …(2)        [Exterior angle property of a triangle]
From (1) and (2),
we have ∠TQR + ½∠P = ∠TQR + ∠T
⇒ ½∠P = ∠T
⇒ ½∠QPR = ∠QTR or ∠QTR = ½∠QPR