What are the measures of the apex angles of all isosceles triangles with circumradius 9 and inradius 4?

(As a reminder, the circumradius is the radius of the circumscribed circle, and the inradius is the radius of the inscribed circle.)

Defining the apex angle, A, as the angle adjacent to the two equal sides of an Isoceles triangle, the apex angle can be given by the equation:
                A = 2*sin^-1(r / (4 * R * sin²(½ B))),
where r is the triangle's inradius, R is the triangle's circumradius, and B is the angle of one of the two internal angles.

Where the inradius is 4 and the circumradius is 9, the apex angle is then
                 2*sin^-1(1 / (9 * sin²(½ B))).

By iteration (using Excel) I find that there are two such angles for the apex. They (approximately) are:
38.9424413 degrees (0.67967397 radians) and
83.6206298 degrees (1.45945531 radians)

Edited on July 6, 2006, 5:21 am

Posted by Dej Mar on 2006-07-05 19:45:07