NCERT Class 9 Solutions: Line and Angles (Chapter 6) Exercise 6.2 – Part 2

Angles on the same side of transversal

Angles on the Same Side of Transversal

Angles on the same side of transversal

Q-3 In the figure, if AB CD , EFCD and GED=126° , find AGE,GEF and FGE .

Angle α Angle α: Angle between E_1, E, E' Angle α Angle α: Angle between E_1, E, E' Segment f Segment f: Segment [F, E] Segment g Segment g: Segment [E, G] Vector u Vector u: Vector[A_1, B_1] Vector u Vector u: Vector[A_1, B_1] Vector v Vector v: Vector[A_1, C_1] Vector v Vector v: Vector[A_1, C_1] Vector w Vector w: Vector[D_1, E_1] Vector w Vector w: Vector[D_1, E_1] Vector a Vector a: Vector[D_1, F_1] Vector a Vector a: Vector[D_1, F_1] Point F Point F: Point on u Point F Point F: Point on u Point F Point F: Point on u Point E Point E: Point on w Point E Point E: Point on w Point E Point E: Point on w Point G Point G: Point on v Point G Point G: Point on v Point G Point G: Point on v Point A Point A: Point on v Point A Point A: Point on v Point A Point A: Point on v Point B Point B: Point on u Point B Point B: Point on u Point B Point B: Point on u Point D Point D: Point on w Point D Point D: Point on w Point D Point D: Point on w Point C Point C: Point on a Point C Point C: Point on a Point C Point C: Point on a

Point of a, B, C, D, E, F, G and Line AB and CD at Parallel

Point of A, B, C, D, E, F, G and line AB and CD at parallel

Solution:

Given AB CD , EFCD and GED=126°

Now,

FED=90°(EFCD)

Now, AGE=GED (Since ,AB||CD and GE is transversal. Alternate interior angles) AGE=126° (GED=126°)

Also,

GEF=GEDFED

GEF=126°90° GEF=36 °

Now, FGE+AGE=180° (Linear pair) FGE=180°126° FGE=54°

Q-4 In the figure, if PQ||ST , PQR=110° and RST=130°, find QRS.

Angle α Angle α: Angle between P, Q, R Angle α Angle α: Angle between P, Q, R Angle α Angle α: Angle between P, Q, R Angle β Angle β: Angle between R, S, T Angle β Angle β: Angle between R, S, T Angle β Angle β: Angle between R, S, T Segment f Segment f: Segment [S, R] Segment g Segment g: Segment [R, Q] Vector u Vector u: Vector[S, B] Vector u Vector u: Vector[S, B] Vector v Vector v: Vector[Q, E] Vector v Vector v: Vector[Q, E] Point S S = (0.62, 2.82) Point S S = (0.62, 2.82) Point S S = (0.62, 2.82) Point R R = (-0.44, 0.78) Point R R = (-0.44, 0.78) Point R R = (-0.44, 0.78) Point Q Q = (-1.98, 2) Point Q Q = (-1.98, 2) Point Q Q = (-1.98, 2) Point P Point P: Point on v Point P Point P: Point on v Point P Point P: Point on v Point T Point T: Point on u Point T Point T: Point on u Point T Point T: Point on u

Point P, Q, R, S, T Also PQ Parallel to ST

Point P, Q, R, S, T also PQ parallel to ST

Solution:

Given,

PQ||ST,PQR=110°andRST=130°

A line XY parallel to PQ and ST is drawn.

Angle α Angle α: Angle between P, Q, R Angle α Angle α: Angle between P, Q, R Angle α Angle α: Angle between P, Q, R Angle β Angle β: Angle between R, S, T Angle β Angle β: Angle between R, S, T Angle β Angle β: Angle between R, S, T Segment f Segment f: Segment [S, R] Segment g Segment g: Segment [R, Q] Segment h Segment h: Segment [A, C] Vector u Vector u: Vector[S, B] Vector u Vector u: Vector[S, B] Vector v Vector v: Vector[Q, E] Vector v Vector v: Vector[Q, E] Point S S = (0.62, 2.82) Point S S = (0.62, 2.82) Point S S = (0.62, 2.82) Point R R = (-0.44, 0.78) Point R R = (-0.44, 0.78) Point R R = (-0.44, 0.78) Point Q Q = (-1.98, 2) Point Q Q = (-1.98, 2) Point Q Q = (-1.98, 2) Point P Point P: Point on v Point P Point P: Point on v Point P Point P: Point on v Point T Point T: Point on u Point T Point T: Point on u Point T Point T: Point on u X text1 = "X" Y text2 = "Y"

Line XY Parallel to PQ and ST

Line XY parallel to PQ and ST

PQR+QRX=180° (Angles on the same side of transversal) 110°+QRX=180°

QRX=70°

Also,

RST+SRY=180° (Angles on the same side of transversal) 130°+SRY=180°

SRY=50°

Now,

QRX+SRY+QRS=180°

70°+50°+QRS=180°

QRS=60°

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