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 Equation are produced to points S and T respectively. If Equation and Equation , find Equation .

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, Equation sides QP and RQ

Equation Now,

Equation (SQ is a straight line) Equation

Also,

Equation (TR is a straight line) Equation

Now,

Equation (Sum of the interior angles of the triangle) Equation

Q-2 In the figure, Equation If YO and ZO are the bisectors of Equation and Equation respectively of Equation , find Equation and Equation

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, Equation

YO and ZO are the bisectors of Equation and Equation respectively.

Now,

Equation (Sum of the interior angles of the triangle) Equation

Now,

Equation (ZO is the bisector) Equation

Also,

Equation (YO is the bisector) Equation

Now,

Equation (Sum of the interior angles of the triangle) Equation

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