This topic examines the application of sine and cosine theorems. Because cosines are known, we can also find out their sine values, sinA=4/5, sinB= 12/ 13, sine theorem: b/sinB=a/sinA, and substitute the data of b to find out a= 13/5. + b？ - a？ ) /(2bc), that is, 3/5=(c? +9- 169/25)/(2x3xc) is equivalent to solving a quadratic equation with one variable. It's quite easy to solve. It's turned into 25c? -90c+56=0

The solution is c 1= 14/5 or c2=4/5. Now there are two results of C, which may not be satisfied, because under the condition that both sides a= 13/5 and b=3 are determined, C is actually restricted, and there cannot be two. Our cosB has not been used yet. + c？ -B? ) /(2ac), we find that only c= 14/5 is satisfied, and 4/5 is not.

So c= 14/5.

There is another way to use angles, which has already been used. I won't use his method, just remember that there are many solutions to this kind of problem, and the simplicity is different.