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

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Figure of Interior and Exterior angle

Figure of Interior and Exterior Angle

Figure of Interior and Exterior angle

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 .

Triangle poly1 Triangle poly1: Polygon R, P, Q Angle α Angle α: Angle between P, Q, T Angle α Angle α: Angle between P, Q, T Angle α Angle α: Angle between P, Q, T Angle β Angle β: Angle between S, P, R Angle β Angle β: Angle between S, P, R Angle β Angle β: Angle between S, P, R Segment d Segment d: Segment [R, P] of Triangle poly1 Segment a Segment a: Segment [P, Q] of Triangle poly1 Segment c Segment c: Segment [Q, R] of Triangle poly1 Vector u Vector u: Vector[R, B] Vector u Vector u: Vector[R, B] Vector v Vector v: Vector[P, E] Vector v Vector v: Vector[P, E] Point R R = (3.4, 0.26) Point R R = (3.4, 0.26) Point R R = (3.4, 0.26) Point P P = (1.96, 3.66) Point P P = (1.96, 3.66) Point P P = (1.96, 3.66) Point Q Point Q: Point on u Point Q Point Q: Point on u Point Q Point Q: Point on u Point T Point T: Point on u Point T Point T: Point on u Point T Point T: Point on u Point S Point S: Point on v Point S Point S: Point on v Point S Point S: Point on v

Triangle PQR Also Side QP and RQ

Triangle PQR also side QP and RQ are produce to point S and T respectively also angle SPR = 135(degree) and angle PQT = 110(degree)

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.

Triangle poly1 Triangle poly1: Polygon R, P, Q Angle α Angle α: Angle between P, Q, T Angle α Angle α: Angle between P, Q, T Angle α Angle α: Angle between P, Q, T Angle β Angle β: Angle between S, P, R Angle β Angle β: Angle between S, P, R Angle β Angle β: Angle between S, P, R Segment d Segment d: Segment [R, P] of Triangle poly1 Segment a Segment a: Segment [P, Q] of Triangle poly1 Segment c Segment c: Segment [Q, R] of Triangle poly1 Vector u Vector u: Vector[R, B] Vector u Vector u: Vector[R, B] Vector v Vector v: Vector[P, E] Vector v Vector v: Vector[P, E] Point R R = (3.4, 0.26) Point R R = (3.4, 0.26) Point R R = (3.4, 0.26) Point P P = (1.96, 3.66) Point P P = (1.96, 3.66) Point P P = (1.96, 3.66) Point Q Point Q: Point on u Point Q Point Q: Point on u Point Q Point Q: Point on u Point T Point T: Point on u Point T Point T: Point on u Point T Point T: Point on u Point S Point S: Point on v Point S Point S: Point on v Point S Point S: Point on v

Triangle of XYZ

Rectangle of XYZ, angle XYZ = 54(degree) if YO and ZO are the bisector of angle XYZ and angle XZY respectively of triangle XYZ

Solution:

Given, X=62°,XYZ=54°

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

Now,

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

Now,

OZY=12XZY (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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