sinA= 3/5 and C=17

Question

Finding a and b

Answer 1

sin A = 3/5 = a/c

if side a = 3 and side c = 5
you should recognize this as a
special rt triangle of 3-4-5
so b = 4

If not
a^2 + b^2 = c^2
3^2 + b^2 = 5^2
b^2 = 16
b = 4

so, a = 3, b = 4