In figure, two line segments `AC` and `BD` intersects each other at the point `P` such that `PA` = 6 cm, `PB` = 3 cm, `PC` = 2.5 cm, `PD`=5 cm, `angleAPB=50^(@)` and `angleCDP=30^(@)`. Then, `anglePBA` is equal to

Options

A. `50^(@)`
B. `30^(@)`
C. `60^(@)`
D. `100^(@)`

Answer

D

Explanation

In `DeltaAPB and DeltaCPD angleAPB=angleCPD=50^(@)` [vertically opposite angle]
`(AP)/(PD)=6/5`….(i)
and `(BP)/(CP)=(3)/(2.5)=6/5….(ii)`
From Eqs. (i) and (ii)
`(AP)/(PD)=(BP)/(CP)`
`thereforeDeltaAPB~DeltaDPC` [by SAS similarity critrion]
`thereforeangleA=angleD=30^(@)`[corresponding angles of similar triangles]
In `DeltaAPB, angleA+angleB+angleAPB=180^(@)` [sum of angles of a triangle=`180^(@)`]
`rArr30^(@)+angleB+50^(@)=180^(@)`
`thereforeangleB=180^(@)-(50^(@)+30^(@))=100^(@)`
i.e., `anglePBA=100^(@)`