In a triangle ABC, D is the midpoint of BC. Join AD. Angle ADB = 45 degree and angle ACB = 30 degree. Find angle ABC.
let AH be a perpendicular line (H belongs to BC)
In the AHC triangle we have AH=AC * 1/2 and HC=AC*(¡î3)/2
HC=AH*¡î3
In the ADH trinangle we have DH=AH
DH+HC=a (a=BC/2) (AH=h)
tg(ABC)=AH/BH = h/a+h = h/h(2+¡î3) ==>
(ABC)=arctg (1/2+¡î3) .......ABC = 105¨¬
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Posted by vije
on 2006-02-26 11:24:11 |
alternative aproach
