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

When a parallel line is cut by transversal cooresponding angles are equal, that is A=E,C=G,B=f,D=H

Cooresponding Angles Are Equal

When a parallel line is cut by transversal cooresponding angles are equal, that is A=E,C=G,B=f,D=H

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=½(QOSPOS).

give point of O,P,Q,R,S and ∠ROQ=90 and also POQ is a line.Ray OR is perpendicular to line PQ.OS is another ray laying between rays OP and OR

Give Point of O,P,Q,R,S and ∠ROQ=90

give point of O,P,Q,R,S and ∠ROQ=90 and also POQ is a line.Ray OR is perpendicular to line PQ.OS is another ray laying between rays OP and OR

Solution,

Given,

  • Points O, P, Q, R, S.

  • POQ is a line.

  • Ray OR is perpendicular to line PQ.

  • OS is another ray lying between rays OP and OR.

To Prove,

ROS=12(QOSPOS).

We know that,

  • 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°+ROQ

  • QOSPOS=2ROQ

  • ROS=1/2(QOSPOS)

XYZ=64°ZYPXYQQYP.

Solution:

Given,

  • Points P, Q, X, Y, Z and XYZ=64°

  • YQ bisects ZYP

P, Q, X, Y, Z and angle XYZ = 64 degrees

XY Produced to P and Angle XYZ Is 64 Degrees

P, Q, X, Y, Z and angle XYZ = 64 degrees

We have,

  • XYZ+ZYP=180° (Linear Pair)

  • 64°+ZYP=180°

  • ZYP=116°

Also,

  • ZYP=ZYQ+QYP

  • ZYQ=QYP (YQ bisects ZYP )

Again,

  • 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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