There is a point M inside a square ABCD such that angle MAB is 60° and angle MCD is 15°. Find angle MBC.
Let point M have coordinates (x,y) and the square have unit sides, with D at the origin and B at (1,1) and A at (0,1).
Since MAB is 60 degrees, MAD is 30 degrees.
So x = (1-y)tan 30
Since angle MCD is 15 degrees,
y = (1-x)tan 15
Substituting this expression for y in the first equation,
x = (1 - tan15 + x tan 15) tan 30
This solves to
x = (1-tan15)tan30/(1-(tan15)(tan30))
This latter comes out to x = .5, so M is on the vertical centerline of the square.
By symmetry then, angle MBC is the same as angle MAD, or 30 degrees.
Edited on June 11, 2004, 9:37 am
Posted by Charlie
on 2004-06-11 09:35:30