The sum of two angles of a triangle is `80^@` and their difference is `20^@` . Find all the angles.

Explanation

Given:
Sum to two angles of a triangle is `80^@` and their difference is `20^@`.
Let the two angles be `/_A`,`/_B`and the third angle be `/_C`.
`/_A+/_B=80^@`
`/_A=80^@-/_B`............(i)
`/_A-/_B=20^@`
`/_A=20^@+/_B`............(ii)
From (i) and (ii), we get:
`80^@-/_B=/_20^@+/_B`
`2/_B=60^@`
`/_B=30^@`
Putting the value of `/_B` in (i)
`/_A+30^@=80^@`
`/_A=50^@`
Now from angle sum of property, we know that
`/_A+/_B+/_C=180^@`
`50^@+30^@+/_C=180^@`
`/_C=180^@-100^@`
`/_C=80^@`.
Hence, the all three angles are `50^@`, `30^@` and `100^@`.