**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:

∠RQT = ∠PQT + ∠PQR *[Linear pair axiom]*

∠PQR = 180° – ∠PQT

= 180° – 110°

∠PQR = 70°

∠SPQ = ∠SPR + ∠RPQ *[Linear pair axiom]*

∠RPQ = 180° – ∠SPR

= 180° – 135°

∠RPQ = 45°

∠PQR + ∠RPQ + ∠PRQ = 180° *[Sum of interior angles of a triangle is 180°]*

∠PRQ = 180° – 45° – 70°

**∠PRQ = 65° **

**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.**

Solution:

Given ∠x = 62° and ∠XYZ = 54°. If YO and ZO are the bisectors of ∠ XYZ and ∠ XZY respectively of ∆ XYZ. We have to find ∠OZY and ∠YOZ

In ∆ XYZ, ∠XYZ +∠XZY + ∠YXZ = 180° *[Sum of interior angles = 180°]*

62° + 54° + ∠XZY = 180°

∠XZY = 180° – 62° – 54° = 64°

Since ZO is the bisectors of ∠XZY in ∆ XYZ, then ∠ OZY = ∠ XZO

∠XZY = ∠OZY + ∠XZO

= ∠OZY + ∠OZY

= 2 ∠OZY

64° = 2 ∠OZY

∠OZY = ^{64°}/_{2 }= 32°

Since YO is the bisectors of ∠ XYZ in ∆ XYZ, then ∠ XYO = ∠ OYZ

∠ XYZ = ∠ XYO + ∠ OYZ

= ∠ OYZ + ∠ OYZ

= 2 ∠OYZ

54° = 2 ∠OYZ

∠OYZ = ^{54°}/_{2 }= 27°

∠OZY + ∠OYZ + ∠YOZ = 180°

32° + 27° + ∠YOZ = 180°

∠YOZ = 180° – 32° – 27° = 121°

Hence ∠YOZ = 121°, ∠OZY _{ }= 32°

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

Solution:

Let AB||DE , ∠ BAC = 35° and ∠ CDE = 53°

∠ BAC = ∠CED = 35° *[alternate angles]*

∠CED +∠CDE + ∠ECD = 180°

∠ECD = 180° – 53° – 35° = 92°

∠ECD = 92°

**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 PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠ TSQ = 75°,we have to find ∠SQT

∠ RPT + ∠ PRT + ∠ RTP = 180° *[Sukm of interior angles is 180°]*

∠ RTP = 180° – 95° – 40° = 45°

∠ STQ = ∠ RTP = 45° *[vertically opposite angles]*

In ΔSTQ , ∠ STQ + ∠ TSQ + ∠ SQT = 180°

75° + 45° + ∠ SQT = 180°

∠ SQT = 60°

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

Solution:

PQ ⊥ PS, PQ || SR, ∠ SQR = 28° and ∠ QRT = 65°

∠PQS + ∠SQR = ∠QST

x + 28° = 65°

x = 65° – 28° = 37°

In ΔPQS, ∠QPS + ∠PSQ + ∠PQS = 180° *[Sum of interior angles is 180°]*

90° + 37° + ∠PSQ = 180°

∠PSQ = 53°

Therefore, x = 37° and y = 53°

**6. In Fig. 6.44, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠ PQR and ∠ PRS meet at point T, then prove that ∠QTR = ^{1}/_{2 }∠QPR.**

Solution:

By exterior angle property, ∠PRS = ∠PQR + ∠QPR

Therefore, ^{1}/_{2}∠PRS = ^{1}/_{2}∠PQR + ^{1}/_{2}∠QPR

i.e., ∠TRS = ∠TQR + ^{1}/_{2 }∠QPR……………..(1)

In ∆QTR, by exterior angle property

∠TRS = ∠TQR + ∠QTR ……………….(2)

From (1) and (2)

∠TQR + ∠QTR = ∠TQR + ^{1}/_{2 }∠QPR

∠QTR =^{1}/_{2 }∠QPR

## 2 thoughts on “Lines and Angles – EXERCISE 6.3 – Class IX”

Comments are closed.