## 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