In the given figure, lines AB and CD intersect at a point P such that `anglePAC=45^(@),angleACP=100^(@)andanglePBD=65^(@)."Find "angleCPA,angleDPBandangleBDP`

Explanation

We know that the sum of the angles of triangle is `180^(@)`. In `DeltaACP`, we have
`anglePAC+angleACP+angleCPA=180^(@)`
`implies45^(@)+100^(@)+angleCPA=180^(@)`
`impliesangleCPA=(180^(@)-145^(@))=35^(@)`
`:.angleDPB=angleCPA=35^(@)" "["vertically opp."angles]`
In `DeltaPBD` we have
`angleDPB+anglePBD+angleBDP=180^(@)`
[sum of the angles of a triangle]
`=35^(@)+65^(@)+angleBDP=180^(@)`
`impliesangleBDP=(180^(@)-100^(@))=80^(@)`
Hence, `angleCPA=35^(@),angleDPB=35^(@)andangleBDP=80^(@)`.