The sides of a triangle `ABC` are in the ratio `3:4:5`. If the perimeter of triangle `ABC` is `60,` then its lengths of sides are:

Explanation

Let `A=3x, B=4x, C=5x`
but `A+B+C=180^(@)`
`rArr 3x+4x+5x=180^(2)`
`rArr 12x=180^(@)`
`rArr x=15^(@)`
`therefore A=3 xx 15^(@)=45^(@)`
`B=4 xx 15^(@)=60^(2)`
`C=5 xx 15^(2)=75^(@)`
Now sinA=`sin45^(@)=1/sqrt(2)`
`sinB=sin60^(2)=sqrt(3)/2`
`sinC=sin75^(@)=(sin45^(@)+30^(@))`
`=sin45^(2)cos30^(2)+cos45^(2)sin30^(@)`
`=1/sqrt(2) xx sqrt(3)/2 + 1/sqrt(2)xx 1/2=(sqrt(3)+1)/(2sqrt(2))`
`therefore a:b:c = sinA:sinB:sinC`
`=1/sqrt(2):sqrt(3)/2:(sqrt(3)+1)/(2sqrt(2))`
`=2:sqrt(6):(sqrt(2)+1)`. Ans.