9th Maths 6.1
NCERT Class 9th solution of Exercise 6.2
NCERT Class 9th solution of Exercise 6.3
Exercise 6.1
Q1. In the given figure lines AB and CD intersect at O, if ∠AOC+∠BOE=70° and ∠BOD = 40° then find the ∠BOE and reflex ∠COE.
Sol.∠AOC=∠BOD_______(1) [Vertically opp. angles]
∠AOC=40° ________(2) [ Given ∠BOD=40°]
∠AOC+∠BOE=70°____(3) [Given]
40°+∠BOE=70° [ from eq. (2) ]
∠BOE=70°-40°
∠BOE=30°__________(4)
Now
∠AOC+∠COE+∠BOE=180° [ By Linear pair]
40°+∠COE+30°=180° [ from eq. (2) & (4) ]
∠COE=180°-70°
∠COE=110°
Reflex∠COE=360°-∠COE
Reflex∠COE=360°-110°
Reflex∠COE=250°
Answer :
The required ∠BOE=30° and reflex∠COE=250°.
Q2. In the figure lines, XY and MN intersect at O. If ∠POY=90° and a:b=2:3, find c.
Sol.∠XOP+∠POY=180° [ By linear pair ]
(∠XOM+∠MOP)+90°=180° [ Given ∠POY=90°]
(a+b)=180°-90°
a+b=90°
Sum of the ratio
a+b=2+3
a+b=5
a=frac(2)(5)times90°
a=36°
b=frac(3)(5)times90°
b=54°_____________(1)
c+b=180° [ by linear pair ]
c+54=180° [ from eq. (1) ]
c=180°-54°
c=126°
Answer :
The required value of c=126°.
∠PQS+∠PQR=180° [ By linear pair ]
∠PQR=180°-∠PQS________(1)
∠PRQ+∠PRT=180° [ By linear pair ]
∠PRQ=180°-∠PRT________(2)
`∠PQR=∠PRQ [ Given ]
180°-∠PQS=∠PRT-180° [ From eq. (1) & (2) ]
∠PQS=∠PRT
Proved.
Q4. In the figure, if x+y=w+z, then prove that: AOB is a line.
Sol.
x+y+w+z=360°____(1) [By angle sum property of a point]
x+y=w+z ____(2) [ Given ]
x+y+x+y=360°
2x+2y=360°
2(x+y)=360°
x+y=frac(360°)(2)
x+y=180°
∠AOB=180°
AOB is a line.
Proved.
Q5. In the figure, POQ is a straight line. Ray OR is perpendicular to line PQ. OS is another ray lying between ray OP. Prove that:
∠ROS=frac(1)(2)(∠QOS-∠POS).
Sol.
∠POR+∠ROQ=180° [ By linear pair ]
∠POR+90°=180° [ Given OR⊥PQ ]
∠POR=90°
So
∠POR=∠ROQ_________(1)
∠POR=∠POS+∠ROS___(2)
∠ROQ=∠QOS-∠ROS___(3)
∠POS+∠ROS=∠QOS-∠ROS[ From eq. (1),(2) & (3) ]
∠ROS+∠ROS=∠QOS-∠POS
2∠RPS=∠QOS-∠POS
∠ROS=frac(1)(2)(∠QOS-∠POS)
Proved.
Q6. It is given that ∠XYZ=64° and XY is produced to point P. Draw a figure from the given information. If ray ∠YO bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Sol.
∠ZYP+∠XYZ=180° [ By linear pair ]
∠ZYP+64°=180° [ Given ∠XYZ=64°]
∠ZYP=180°-64°
∠ZYP=116°
∠ZYQ+∠QYP=116°
∠ZYQ+∠ZYQ=116° [Given Ray YQ bisect ∠ZYP ]
2∠ZYQ=116°
∠ZYQ=frac(116°)(2)
∠ZYQ=58°=∠QYP
∠XYQ=∠XYZ+∠ZYQ
∠XYQ=64°+58° [ ∠XYZ=64°]
∠XYQ=122°
Reflex∠QYP=360°-∠QYP
Reflex∠QYP=360°-58°
Reflex∠QYP=302°
Answer :
The required ∠XYQ=122° and Reflex ∠QYP=302°.
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