Solve the triangle in which sin A :sin B :Sin C =3:4:6and it's perimeter equals 52 cm
Let the sides of triangle ABC be a, b, and c ,so that
a is opposite angle A, etc
we know a+b+c = 52, so c = 52-a-b
sinA/sinB = a/b
a/b = 3/4
4a = 3b---> b = 4a/3
sinA/sinC = a/c
3/6 = a/(52-b-a)
1/2 = a/(52-a-b)
2a = 52 - a - b
3a = 52-b
3a = 52 - 4a/3
9a = 156 - 4a
13a = 156
a = 12
then b = 4(12)/3 = 16
c = 52-12-16 = 24
notice that 12 : 16 : 24 = 3 : 4 : 6 !
( I actually did not know that the ratio of sides is equal to the ratio of the sines of corresponding angles, you learn something new even when you are quite old)
Now that we know the lengths of the sides, we can use the cosine law to find one of the angles, I suggest finding C,
then the sine law to find one of the others,
then using the fact that all three add up to 180 to find the third.