9th Mahts 6.3
NCERT Class 9th solution of Exercise 6.1
NCERT Class 9th solution of Exercise 6.2
Exercise 6.3
Q1. In the given figure, sides QP and RQ of △PQR are produced to points S and T respectively. If ∠SPR=135° and ∠PQT=110°, find ∠PRQ.
Sol.
∠PQT+∠PQR=180° [ By linear pair. ]
110°+∠PQR=180° [ Given ∠PQT=110° ]
∠PQR=180°-110°
∠PQR=70°
∠PRQ+∠PQR=∠SPR
[ By exterior angle property ]
∠PRQ+70°=135° [ Given ∠SPR=135° ]
∠PRQ=135°-70°
∠PRQ=65°
Answer :
The required value of ∠PRQ=65°.
Q2. In the given figure, ∠X=62°, ∠XYZ=54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of △XYZ, find ∠OZY and ∠YOZ.
Sol. :
In △XYZ,
∠YXZ+∠XYZ+∠XZY=180°
[ By Angle sum property of △]
62°+54°+∠XZY=180°
∠XZY=180°-62°-54°
∠XZY=180°-116°
∠XZY=64°
∠XZO=∠OZY=frac(1)(2)times∠XZY
[ ZO is bisect ∠XZY]
∠XZO=∠OZY=frac(1)(2)times64°
[ Given ∠XZY=64°]
∠XZO=∠OZY=32°
∠XYO=∠OYZ=frac(1)(2)times∠XYZ
[ YO is bisect ∠XYZ ]
∠XYO=∠OYZ=frac(1)(2)times54°
[ Given ∠XYZ=54° ]
∠XYO=∠OYZ=27°
In △YOZ
∠OYZ+∠OYZ+∠YOZ=180°
[ By angle sum property of △]
27°+32°+∠YOZ=180°
∠YOZ=180°-27°-32°
∠YOZ=180°-59°
∠YOZ=121°
Answer :
The required ∠OZY=32° and ∠YOZ=121°.
Q3. In the given figure, if AB∥DE, ∠BAC=35° and ∠CDE=53°, find ∠DCE
Sol. :
AB∥DE and AE is transversal. [ Given ]
∠CED=∠BAC [ Alternate angles ]
∠CED=35°
In △CDE
∠DCE+∠CDE+∠CED=180°
[ By angle sum property of △ ]
∠DCE+53°+35°=180°
[Given ∠CDE=53°]
∠DCE=180°-88°
∠DCE=92°
Answer :
The required value ∠DCE=92°.
Sol. :
In △PRT
∠PTR+∠RPT+∠PRT=180°
[ By angle sum property of △ ]
∠PRT+95°+40°=180°
[ Given ∠RPT=95° & ∠PRT=40°]
∠PRT=180°-95°-40°
∠PRT=180°-135°
∠PRT=45°
PQ intersect RS at point T
∠QTS=∠PRT [ Vertically opp. angles]
∠QTS=45°
In △SQT
∠SQT+∠TSQ+∠QTS=180°
[ By angle sum property of △]
∠SQT+75°+45°=180°
[ Given ∠TSQ=75°]
∠SQT=180°-75°-45°
∠SQT=180°-120°
∠SQT=60°
Answer :
The required value of ∠SQT=60°.
Sol. :
PQ∥SR and QR is transversal
∠PQR=∠QRT [Alternat angles ]
According to figure
∠PQS+∠SQR=∠QRT
x+28°=65°
[Given ∠SQR=28° & ∠QRT=65°]
x=65°-28°
x=37°
In △PQS
∠QPS+∠PQS+∠PSQ=180
[ By angle sum property of △]
90°+37°+y=180° [ Given ∠QPS=90°]
y=180°-90°-37°
y=180°-127°
y=53°
Answer :
The required value of x=37° & y=53°.
According to figure
∠PRS=∠QPS+∠PQR______(1)
[ By exterior angle property ]
∠TRS=∠QTR+∠TQR_____(2)
[ By exterior angle property ]
Since QT & RT are bisectors of ∠PQR
& ∠PRS [Given]
∠TQR=frac(1)(2)∠PQR___(3)
∠TRS=frac(1)(2)∠PRS____(4)
∠TRS=frac(1)(2)∠QPR+frac(1)(2)∠PQR___(5)
[ by eq.(1) &(4)]
∠QTR+∠TQR=frac(1)(2)∠QPR+frac(1)(2)∠PQR
[ by eq.(2) & (5) ]
∠QTR+frac(1)(2)∠PQR=frac(1)(2)∠QPR+frac(1)(2)∠PQR
∠QTR=frac(1)(2)∠QPR
Proved.
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