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Question 5
In figure, two line segments AC and BD intersect each other at the point P such that PA=6cm,PB=6cm,PC=3cm,PC=2.5cm,PD=5cm,APB=50 and CDP=30 .Then PBA is equal to




(A) 50
(B) 30
(C) 60
(D) 100

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Solution

(D) 100

In ΔAPB and ΔCPD,
APB=CPD=50 [ Vertically opposite angles ]
APPD=65............(i)
BPCP=32.5=65.............(ii)
From Eq.s (i) and (i)
APPD=BPCP
ΔAPBΔDPC [ by SAS similarity criterion ]
A=D=30 [ corresponding angles of similar triangles ]
In ΔAPB ,
A+B+APB=180 [ sum of angles of a triangle = 180 ]
30+B+50=180
B=180(50+30)=100
i.e., PBA=100

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