There is a point M inside a square ABCD such that angle MAB is 60° and angle MCD is 15°. Find angle MBC.

(In reply to Generalization by Nick Hobson)

Yes. Putting the point M at (x,y) in the unit square [0,1]x[0,1], the given angles have tangents y/x and (1-y)/(1-x) and the tangent of the sought angle is (1-x)/y. Now

((1-x)/y)*(1-(1-y)/(1-x))=((1-x)/y) - ((1-y)/y)=1-x/y  

and your formula is verified. When the given angles are both 45 degrees, your formula gives an indeterminate 0/0 and, indeed, the position of M can be anywhere on the line AC so the sought angle can be anything from 0 to 90 degrees.

Edited on June 15, 2004, 3:10 am

Posted by Richard on 2004-06-14 19:33:34