ABC, AB = BC = 25 and AC = 40. What is sin angle ACB?
Triangle ABC is isosceles. Picture a line perpendicular to AC and half way along, going from AC to B . It will have a height given by sqrt(252 - 202) by Pythagoras. Hence the sin(ACB) = sqrt(252 - 202)/25 = sqrt(5*45)/25 = 15/25 = 3/5
Let's name the angle corresponding to one of the isosceles sides "x". With this, we have 2 base angles of "x" degrees(on the side of length 40), and a third angle of measurement 180-2x degrees. By dropping an altitude to side AC, we form two right triangles, both of which are congruent to each other AAS. Since the problem asks for the sin of ACB, we need only the opposite side of the right triangle, which is the altitude, divided by the hypotenuse. Because the two right triangles formed are congruent, the base of 40 is split into two halves of length 20. We then realize that this forms a 3 - 4 -5 triangle with a scale factor of 5:
Meaning that the length of the altitude of AC is 15 (3 * 5). Now that we have the opposite side, sine is just opposite / hypotenuse, which is 15 / 25 giving us our final answer of 3/5