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)
Let's assume that the triangle MAB is an equalateral triangle, then fill in the values for all the angles made by the additional lines drawn.
Angel MAB is 60°, so angle MAD is 30°.
If angle MCD is 15°, then because we are assuming an equalateral triangle at MAB, then angle MDC is also 15°.
If angle MDC is 15°, then angle MDA is 75°.
If angle MAD is 30° and angle MDA is 75°, then angle AMD is also 75°.
Filling in the mirror image side...
ABM = 60°, MBC = 30°, BCM = 75°, and BMC = 75°.
Still assuming that triangle ABM is equalateral, we claim that angle AMB is also 60°.
Adding up all the known angles around point M (75°+60°+75°=210°) and subtracting it from 360° we get a value for angle CMD of 150°.
Now all we have to do is verify that the angles we provided for triangle CMD add up to 180°... 15°+15°+150°=180° so that works.
Posted by Erik O.
on 2004-06-11 09:51:39