Question

Question #b5018

Answer 1

`angleAPB=80^@`

Explanation:

As shown in the diagram, `ABCD` is a irregular quadrilateral.

One common property of all quadrilaterals is that the sum of all their angles equals `360^@`
`=> anglea+angleb+anglec+angled=360^@`

Given that `anglec=100^@, angled=60^@`, and `BP and AP` bisects `angleb and anglea`, respectively,

`=> anglea+angleb=360-(anglec+angled)=360-(100+60)=200^@`
`=> (anglea)/2+(angleb)/2=200/2=100^@`

`=> angleAPB=180-((anglea)/2+(angleb)/2)=180-100=80^@`