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 45-45-90 and 30-60-90 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=[2-sqrt(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 2006-02-26 03:50:20