In a triangle ABC, D is the midpoint of BC. Join AD. Angle ADB = 45 degree and angle ACB = 30 degree. Find angle ABC.
A similar approach to that of vije...
Extend CB to O, so that AO is perpendicular to CO.
WLOG, letting CD = DB = 1,
cot 30° = (2 + BO)/AO = sqrt(3),
cot 45° = (1 + BO)/AO = 1.
Hence BO/AO = cot ABO = 2 - sqrt(3).
(The three terms form an arithmetic progression.)
Now we can either note(!) that cot 75° = 2 - sqrt(3), or note that we can use the addition formula for cotangents:
cot 75° = cot (45° + 30°)
= (cot 45° cot 30° - 1)/(cot 45° + cot 30°)
= (sqrt(3) - 1)/(1 + sqrt(3))
= 2 - sqrt(3)
Either way, angle ABO = 75°, and so angle ABC = 105°.
Edited on February 26, 2006, 1:00 pm
Another approach
