In triangle ABC, angle C is 90 degrees AC is equal to 27 sinA = 3/5 find BC.

An error was made in the problem statement. If the sine of angle A were equal to three-fifths, then it would be impossible to get BC equal to 36 (and in general the task would be impossible) Instead of the sine there should be a cosine equal to 3/5. Then:
1) We draw a triangle ABC with a right angle C.
2) Note in the figure that AC = 27
3) cosA = AC / AB AC / AB = 3/5 27 / AB = 3/5 AB = 45
4) Now we know the hypotenuse and leg, so it won’t be difficult to find the second leg. You can notice that in our triangle there are Pythagorean numbers: 27 (3 * 9), 36 (4 * 9), 45 (5 * 9).
p.s For those who did not see that there are Pythagorean numbers here, they can count according to the Pythagorean theorem: c ^ 2 = a ^ 2 + b ^ 2