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)]/[1sqrt(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 20060225 15:15:52 