In a triangle ABC AC = BC = 10, AB = 8√6. Find sin A.

Let’s build the height of the CH to the AB side.

Since, by condition, AC = BC, the triangle ABC is isosceles, and then its height CH is also the median and bisector.

Then AH = BH = AB / 2 = 8 * √6 / 2 = 4 * √6 cm, and the triangles ACH and BCH are rectangular.

In a right-angled triangle ACH, according to the Pythagorean theorem, we determine the length of the leg CH.

CH ^ 2 = AC ^ 2 – AH ^ 2 = 100 – 96 = 4.

CH = 2 cm.

SinCAB = CH / AC = 4/10 = 2/5.

Answer: The sine of angle A is 2/5.