Originally Posted by

**Soroban** Hello, Amy!

Here's another approach for finding the length of chord AB.

Let O = the vertex of the isosceles triangle.

. . and AB = the base of the triangle.

The vertex angle is: 3π/4N. .OA = OB = 12.

The Law of Cosines:

. . . . . . . .ABČ .= .OAČ + OBČ - 2(OA)(OB)cosθ

We have: .ABČ .= .12Č + 12Č - 2(12)(12)cos(3π/4N)

. . . . . . . . . . . = .288 - 288·cos(3π/4N)

. . . . . . . . . . . = .288·[1 - cos(3π/4N)]

. . . . . . . . . . . = .288·[2·sinČ(3π/8N)] . ← . Half-angle identity

. . . . . . . . . . . = .576·sinČ(3π/8N)

Therefore: .AB .= .24·sin(3π/8N)

Of course, Jhevon's right-triangle approach is faster and simpler.

. . As usual, he did a great job.