Question 791763
This happens to be one of the basic right triangles that I remember, 5:12:13, but to calculate it perform the following steps.
Let a = one of the legs of the right triangle
Let b = the other leg
Then we have
(1) a + b = 30 - 13 or
(2) a + b = 17 or
(3) a = 17 - b
Now using pythagoreon theorem we have
(4) {{{a^2 + b^2 = 13^2}}} or
(5) {{{a^2 + b^2 = 169}}}  
Now substitute a of (3) into (5) to get
(6) {{{(17-b)^2 + b^2 = 169}}} or
(7) {{{289-34*b+b^2 + b^2 = 169}}} or
(8) {{{289-169-34*b+2*b^2 = 0}}}  or using x = b
(9) {{{2*x^2-34*x+120 = 0}}} 
Use the quadratic equation to solve for x
(10) {{{x = (34 +- sqrt( 34^2-4*2*120 ))/(2*2) }}}  
(11)  b = 5,12
Let's check this with (1) and (4).
Is (5+12+13 = 30)?
Is (30 = 30)? Yes
Is ({{{13^2 = 5^2 + 12^2}}})?
Is (169 = 25 + 144)?
Is (169 = 169)? Yes
Answer: the other two sides (legs) are 5cm and 12cm.