In a triangle ABC AC = BC = 20, AB = 32. find sin A.

To solve the problem, let us draw the height from the vertex C to the point H. Since, according to the condition of the problem, AC = BC, the triangle is isosceles. Therefore, according to the theorem: in an isosceles triangle, the height drawn to the base is the bisector and the median.
Therefore, the CH height divides the ВA side in half, that is, BH = AH = AB / 2 = 32/2 = 16.
Since CH is the height, we get two right-angled triangles BCH and ACH.
Consider the ACH triangle: AC – hypotenuse, AH and CH – legs.
The sine of an acute angle in a right triangle is the ratio of the opposite leg to the hypotenuse, in our problem: sin A = a / c = CH / AC.
We calculate the hypotenuse AC using the Pythagorean theorem: in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.
AC ^ 2 = AH ^ 2 + CH ^ 2.
Hence:
CH = √ (AC ^ 2 – AH ^ 2) = √ (20 ^ 2 – 16 ^ 2) = √ (400 – 256) = √144 = 12.
Hence:
sin A = 12/20 = 3/5 = 0.6.

Answer: sin A = 0.6.