In ` A B C ,\ /_B=35^0,/_C=65^0` and the bisector of `/_B A C` meets `B C` in `Pdot` Arrange `A P ,\ B P\ a n d\ C P` in descending order.

Explanation

Given : `angleB=35^@ and angleC=65^@`
Solution : We have to arrange AP, BP and CP in descending order.
In `triangle ACP`
`angle ACP = 65^@`
`angle CAP = 40^@`( As AP is the bisector of `angle CAB` )
So AP > CP ( Sides in front or greater angle will be greater )________1
In ` triangle ABP `
` angleBAP > angleABP `
So BP > AP ______2
Hence, from 1 and 2
BP>AP>CP