NCERT Class 9 Solutions: Line and Angles (Chapter 6) Exercise 6.3 – Part 1

Figure of Interior angle

Give Interior angle also made the triangle

Give Interior Angle

Give Interior angle also made the triangle

Q-1 In the figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If SPR=135° and PQT=110° , find PRQ .

Give ΔPQR also side QP and RQ are produce to point S and T respectively also ∠SPR = 135° and ∠PQT = 110°

Give ΔPQR Also Side QP and RQ

Give ΔPQR also side QP and RQ are produce to point S and T respectively also ∠SPR = 135° and ∠PQT = 110°

Solution:

Given, ΔPQR sides QP and RQ

SPR=135°andPQT=110°

  • Now, SPR+QPR=180° (SQ is a straight line.) 135°+QPR=180°QPR=45°

  • also, PQT+PQR=180° (TR is a straight line.) 110°+PQR=180°PQR=70°

  • Now, PQR+QPR+PRQ=180° (Sum of the interior angles of the triangle.) 70°+45°+PRQ=180°115°+PRQ=180°PRQ=65°

Q-2 In the figure, XYZ=54°. If YO and ZO are the bisectors of XYZ and XZY respectively of ΔXYZ , find OZY and YOZ.

Give reiangle of XYZ, ∠XYZ = 54° if YO and ZO are the bisector of ∠XYZ and ∠XZY respectively of Δ XYZ

Give Reiangle of XYZ

Give reiangle of XYZ, ∠XYZ = 54° if YO and ZO are the bisector of ∠XYZ and ∠XZY respectively of Δ XYZ

Solution:

Given, X=62°,XYZ=54°

YO and ZO are the bisectors of XYZ and XZY respectively.

  • Now,

    X+XYZ+XZY=180° (Sum of the interior angles of the triangle) 62°+54°+XZY=180°116°+XZY=180°XZY=64°

  • Now,∠OZY = 1/2∠XZY (ZO is the bisector.) OZY=12(64°)=32°

  • Also, OYZ=12XYZ (YO is the bisector.) OYZ=12(54°)=27°

  • Now, OZY+OYZ+O=180° (Sum of the interior angles of the triangle.) 32°+27°+O=180°59°+O=180°O=121°

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