NCERT Class 9 Solutions: Line and Angles (Chapter 6) Exercise 6.1 – Part 3

Equality of Corresponding angles

Equality of Corresponding Angles

Equality of Corresponding angles

Q-5 In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ROS=12(QOSPOS).

Angle α Angle α: Angle between Q, O, Q' Angle α Angle α: Angle between Q, O, Q' Vector u Vector u: Vector[O, B] Vector u Vector u: Vector[O, B] Vector v Vector v: Vector[O, C] Vector v Vector v: Vector[O, C] Vector w Vector w: Vector[O, D] Vector w Vector w: Vector[O, D] Vector a Vector a: Vector[O, E] Vector a Vector a: Vector[O, E] Point O O = (0.32, 1.14) Point O O = (0.32, 1.14) Point O O = (0.32, 1.14) Point S Point S: Point on a Point S Point S: Point on a Point S Point S: Point on a Point R Point R: Point on w Point R Point R: Point on w Point R Point R: Point on w Point Q Point Q: Point on u Point Q Point Q: Point on u Point Q Point Q: Point on u Point P Point P: Point on v Point P Point P: Point on v Point P Point P: Point on v

Point of O, P, Q, R, S and Angle ROQ Is 90 Degree

Point of O, P, Q, R, S and POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.

Solution,

Give point O, P, Q, R, S and POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR.

Prove,

ROS=12(QOSPOS)

Equation, POR=ROQ=90° (Perpendicular)

QOS=ROQ+ROS=90°+ROS ……….equation (1) POS=PORROS=90°ROS ………..equation (2)

Subtracting (1) from (2) QOSPOS=(90°+ROQ)(90°ROQ)

QOSPOS=90°+ROQ90°+ROQQOSPOS=2ROQROS=12(QOSPOS)

XYZ=64°ZYPXYQQYP.

Solution:

Give point of P,Q,X,Y,Z and XYZ=64°

YQ bisects ZYP

Angle α Angle α: Angle between Z, Y, X Angle α Angle α: Angle between Z, Y, X Angle α Angle α: Angle between Z, Y, X Segment f Segment f: Segment [X, P] Segment g Segment g: Segment [Y, Q] Segment h Segment h: Segment [Y, Z] Point X X = (-2.22, 1.32) Point X X = (-2.22, 1.32) Point X X = (-2.22, 1.32) Point P P = (3.46, 1.32) Point P P = (3.46, 1.32) Point P P = (3.46, 1.32) Point Y Point Y: Point on f Point Y Point Y: Point on f Point Y Point Y: Point on f Point Q Q = (2.66, 4.14) Point Q Q = (2.66, 4.14) Point Q Q = (2.66, 4.14) Point Z Z = (-2.06, 4.06) Point Z Z = (-2.06, 4.06) Point Z Z = (-2.06, 4.06)

Point of P,Q,X,Y,Z and Angle XYZ 64 Degree

Point of P, Q, X, Y, Z and XYZ 64 degree also XY is produced to point P

XYZ+ZYP=180° (Linear Pair) 64°+ZYP=180° ZYP=116°

Also,

ZYP=ZYQ+QYP ZYQ=QYP (YQ bisects ZYP )

ZYP=ZYQ+ZYQ ZYP=2ZYQ 2ZYQ=116° (ZYP=116°)ZYQ=58°=QYP (ZYQ=QYP)

Now, XYQ=XYZ+ZYQ XYQ=64°+58° (putvalueofXYZandZYQ)

XYQ=122°

Also, reflex QYP=180°+XYQ QYP=180°+122°

QYP=302°

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