In a triangle ABC, D is the midpoint of BC. Join AD. Angle ADB = 45 degree and angle ACB = 30 degree. Find angle ABC.
(In reply to
re(3): Solution  This is how long it took me: by Bractals)
So I will forego the trig, just knowing the ratios of side lengths of 454590 and 306090 triangles.
I begin by constructing a 90 degree angle at a point X, collinear with C, D, and B so that AXB is 90 degrees.
WLOG let CD = DB = 1. CX=2+BX=AX*sqrt(3)=(1+BX)sqrt(3)
So 2+BX=sqrt(3)+BXsqrt(3) or BX=[2sqrt(3)]/[sqrt(3)1]
Multiplying top and bottom by sqrt(3)+1 we have:
BX=[sqrt(3)1]/2 also AX=[sqrt(3)+1]/2
BX^2 + AX^2 is 2, so AB is sqrt(2)
Now we see CB:AB = AB:DB = sqrt(2):1
Therefore ABC is similar to DBA.
It follows from this that ABC is 105 degrees
Edited on February 26, 2006, 3:51 am
Edited on February 26, 2006, 3:53 am
Edited on February 26, 2006, 1:45 pm

Posted by Eric
on 20060226 03:50:20 