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
Solution (unproven) by Jer)
When all else fails, I prove analytically:
Orient the square with A=(0,0) B=(0,1) C=(1,1) D=(1,0)
The line through A has equation y=(sqrt(3)/3)x
The line through C has equation y=(2+sqrt(3))x  (1+sqrt(3))
Solving this system yields (after quite a bit of algebra)
x=sqrt(3)/2 and y=1/2
This implies MA = 1 = AB so triangle MAB is isosceles with vertex A = 60 degrees. The two base angles are then also 60 degrees and since MBA = 60 and ABC = 90, MBC = 30.
Jer

Posted by Jer
on 20040611 09:32:00 