NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Lines and Angles |

Exercise |
Ex 4.3 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 4 Lines and Angles Ex 4.3

**Question 1.**

**In figure, sides QP and RQ of APQR are produced to points S and T, respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.**

**Solution:**

**Question 2.**

**In figure, ∠X – 62°, ∠XYZ = 54°, if YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆XYZ, find ∠OZY and ∠YOZ.**

**Solution:**

In ∆XYZ,

∵ ∠X+ ∠Y+ ∠Z = 180°

(Sum of all angles of triangle is equal to

∴ 62° + ∠Y + ∠Z = 180° [YZX = 62° (Given)]

⇒ ∠Y + ∠Z = 118° .

⇒ \(\frac { 1 }{ 2 }\)∠Y + \(\frac { 1 }{ 2 }\) ∠Z = \(\frac { 1 }{ 2 }\) x 118°

⇒ ∠OYZ + ∠OZY = 59°

(∵ YO and ZO are the bisectors of ∠XYZ and ∠XZY)

**Question 3.**

**In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 539 , find ∠DCE.**

**Solution:**

We have AB || DE

⇒ ∠AED= ∠ BAE (Alternate interior angles)

Now, ∠ BAE = ∠ BAC

⇒ ∠ BAE = 35° [ ∵ ∠ BAC = 35° (Given) ]

∴ ∠ AED = 35°

In ∆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 figure, if lines PQ and RS intersect at point T, such that ∠ PRT = 40°, ∠ RPT = 95° and ∠TSQ = 75°, find ∠ SQT.**

**Solution:**

∵∠PTS = ∠RPT + ∠PRT (Exterior angle = Sum of interior opposite angles)

∠ PTS = 95° + 40° [∵ ∠PPT = 95° (Given)]

⇒ ∠ PTS = 135° [and ∠PRT = 40°]

Also, ∠ TSQ + ∠ SQT = ∠ PTS (Exterior angle = Sum of interior opposite angles)

⇒ 75°+ ∠ SQT = 135°

⇒ ∠ SQT = 60° [∵ ∠ TSQ = 75° (Given)]

**Question 5.**

**In figure, if PQ ⊥ PS, PQ||SR, ∠SQR = 2S° and ∠QRT = 65°, then find the values of x and y.**

**Solution:**

Here, PQ || SR

⇒ ∠ PQR = ∠ OPT (Alternate interior angles)

⇒ x + 28° = 65° ⇒ x = 37°

Now, in right angled ASPQ, we have ∠P = 90°

∴ ∠P + x + y = 180° (∵ Sum of all angles of a triangle is equal to 180°)

⇒ 90°+ 37°+ y= 180°

⇒ 127°+ y=180°

⇒ y = 53°

**Question 6.**

**In figure, the side QR of A PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that**

**Solution:**

In ∆ PQP,

∵ ∠QPR + ∠PQR = ∠PRS …(i)

(∵ Sum of interior opposite angles = Exterior angle)

Now, in ∆ TOR,

∵ ∠QTR + ∠TQR = ∠TRS …..(ii)

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