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

Angles on the same side of transversal, so 3 and 6 are same and 4 and 5 will be same.

Angles on the Same Side of Transversal

Angles on the same side of transversal, so 3 and 6 are same and 4 and 5 will be same.

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

Give point of A,B,C,D,E,F,G and line AB and CD at parallel also EF⊥ CD ,∠GED = 126°

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

Give point of A,B,C,D,E,F,G and line AB and CD at parallel also EF⊥ CD ,∠GED = 126°

Solution:

Given ABCD , EFCD and GED=126°

  • Now, FED=90°(EFCD)

  • Now, AGE=GED (Since ,ABCD 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.

Give the point P,Q,R,S,T also PQ ST and ∠PQR = 110° and ∠RST = 130°

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

Give the point P,Q,R,S,T also PQ ST and ∠PQR = 110° and ∠RST = 130°

Solution:

  • Given,

  • PQST,PQR=110°andRST=130°

  • Construction,

  • A line XY parallel to PQ and ST is drawn.

    Give line XY parallel to PQ and ST also ∠PQR = 110° and ∠RST = 130°

    Give Line XY Parallel to PQ and ST

    Give line XY parallel to PQ and ST also ∠PQR = 110° and ∠RST = 130°

  • 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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