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: Solution - A little jumpy, don't you think? by Eric)

 2 cos(15)         1
   ----------- = ------------
      sin(?)      sin(135-?)

So:  2 cos(15) sin(135-?) = sin(?)  

or  2 cos(15) [sin(135)cos(?)-cos(135)sin(?)]  = sin(?)  

or  sqrt(2) cos(15) [cos(?)+sin(?)]  = sin(?)  

sqrt(2) cos(15) [1+tan(?)]  = tan(?)  

sqrt(2) cos(15)  = tan(?)[1 - sqrt(2) cos(15)]

[sqrt(2) cos(15)]/[1-sqrt(2) cos(15)]   = tan(?)

1/[2sqrt(2)sin(15)-1]   = tan(?)

or ? = ArcTan(1/[2sqrt(2)sin(15)-1]) = -75 or 105

I guess I am wondering if someone could explain what is the intuitive insight that I am missing. I mean obviously cos(15) = sin(105), but I wouldn't have thought to guess that. Perhaps Tan(a+b) identity...

Edited on February 25, 2006, 3:17 pm

Posted by Eric on 2006-02-25 15:15:52