NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Lines and Angles |

Exercise |
Ex 6.3 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles Ex 6.3

**Question 1.**

In the given figure, sides QP and RQ of ΔPQR are produced to points S and T respectively.

If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

**Solution:**

∠RPS + ∠RPQ = 180 °

135° + ∠RPQ = 180° [∵ ∠RPS = 135° (given)]

⇒ RPQ = 180°-135° = 45°

Now, ∠RPQ + ∠PRQ = ∠PQT [ext. angle = sum of int. opp. angles]

⇒ 45° + ∠PRQ = 110°

⇒ ∠PRQ = 110° – 45° = 65°

**Question 2.**

In the given figure, ∠X= 62°,∠XYZ= 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find ∠OZY and ∠YOZ.

**Solution:**

Given, ∠X = 62° and ∠XYZ = 54°

In ΔXYZ, ∠XYZ + ∠YXZ + ∠XZY = 180° [since a sum of all the angles of a triangle is 180°]

54°+62°+∠180°⇒ 116°+∠XYZ =18O°

∠XZY=180°-116°

∴ ∠XZY = 64°

Also, given that YO and ZO are the bisectors of ∠XYZ and ∠XZY, respectively.

So, ∠OYZ = ∠XYZ => ∠OYZ = x 54° = 27°

and ∠OZY = ∠XZY= x 64°= ∠OZY=32°

Now, in ΔOYZ, ∠OYZ + ∠OZY + ∠YOZ = 180° [since, sum of all the angles of a triangle Is 180°]

27° + 32° + ∠YOZ = 180°

59°+∠YOZ = 180° ⇒ ∠YOZ = 180°—59°

∴ ∠YOZ = 121°

Hence,∠OZY = 32° and ∠YOZ = 121°

**Question 3.**

In figure,If AB DE, ∠BAC= 35°and ∠CDE= 53°,find ∠DCE.

**Solution:**

We have, AB || DE

⇒ ∠AED = ∠ BAE (Alternate interior angles)

Now, ∠BAE = ∠BAC

⇒ ∠BAE=35° [ ∵ ∠BAC = 35°Given)]

∠AED = 35°

In Δ DCE,

∴ ∠DCE + ∠CED+ EDC = 180°, (∵ Sum of all angles of triangle is equal to 180°)

⇒ ∠DCE + 35° + 53° – 180° (∵∠AED = ∠CED = 35°)

⇒ ∠DCE = 180 °-(35° + 53° ) ⇒ ∠DCE = 92°

**Question 4.**

In the given figure, if lines PQ and RS intersect at point T such that ∠PRT = 40°, ∠RPT = 95°

and ∠TSQ = 75° then find ∠SQT.

**Solution:**

Given, ∠PRT = 40°, ∠RPT = 95° and

∠TSQ = 75°

In Δ PRT, ∠PRT + ∠RPT + ∠PTR = 180° …(i) [since, sum of all the angles of a triangle is 180°]

On putting ∠PRT =40° and ∠RPT = 95° in eq. (i), we get

40° + 95° + ∠PTR = 180°

⇒ 135° + ∠PTR = 180° ⇒ ∠PTR = 180° -135° ⇒ ∠PTR = 45°

Now, ∠PTR = ∠QTS [vertically opposite angles]

∴ ∠QTS – 45°

In ∠TQS, ∠QTS + ∠TSQ + ∠TQS = 180° … (ii)

[since, sum of all the angles of a triangle is 180°]

On putting ∠QTS = 45° and ∠TSQ = 75° in eq. (ii), we get

45° + 75° + ∠TQS = 180° ⇒120° + ∠TQS = 180°

⇒ ∠TQS – 180°-120°

∴ ∠TQS =60°

Hence,∠TQS = ∠SQT = 60°

**Question 5.**

In the given figure, if PQ ⊥ PS, PQ || SR, ∠SQR – 28° and ∠QRT – 65° then find the values of x and y.

**Solution:**

For Δ QSR, ∠QRT is an exterior angle

∠QRT = ∠SQR + ∠QSR [∵ exterior angle = sum of interior opposite angles]

⇒ 65° = 28° + ∠QSR [∵ ∠QRT = 65° and ∠SQR = 28°, given]

∠QSR = 65° – 28° ⇒ ∠QSR = 37°

Given PQ || SR and SQ is the transversal which intersects PQ and ST at Q and S, respectively.

∴ ∠QSR = ∠PQS [alternate interior angles]

⇒ x = 37°

Now, in ΔPQS, ∠SPQ + ∠PQS + ∠PSQ = 180°[since, sum of all the angles of a triangle is 180°]

⇒ 90° + 37° + y = 180° [∵ PQ ⊥ PS ⇒ ∠SPQ = 90°]

⇒ 127° + y= 180°⇒ y = 180° -127° = 53°

Hence, x = 37° and y= 53°.

**Question 6.**

In the given figure, 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 = ∠QPR.

**Solution:**

In ΔTQR, ∠QTR + ∠TQR + ∠QRT = 180^{0 }[sum of all the angles of a triangle is 180°]

∠QRT = 180^{0 }– [∠TQR + ∠QRT]

∠PQR + ∠QRP + ∠PRT

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