¥Ñ E.T ©ó ¬P´Á¤» ¤T¤ë 29, 2003 2:41 pm
Let ¡çPCA = x , ¡çPCB = 80 ¢X- x
By sine law.
In ¡µPAB, ¡î PA/sin40 ¢X= PB/sin20¢X
¡ï PA = PB*sin40¢X/sin20¢X ---- (a)
In ¡µPBC, ¡î PC/sin10¢X= PB/sin(80¢X- x )
¡ï PC = PB*sin10¢X/sin(80¢X- x) ---- (b)
In ¡µPCA, ¡î PA/sinx = PC/sin30¢X ---- (c)
Sub (a) and (b) into (c).
We have : sin30¢X*sin40¢X/sin20¢X*sin10¢X= sinx/sin(80¢X- x)
sin20¢X*sin10¢X/sin30¢X*sin40¢X= sin(80¢X- x)/sinx
Use compound angle formulae : sin(A-B) = sinA*cosB - cosA*sinB
sin20¢X*sin10¢X/sin30¢X*sin40¢X= sin80¢X*cosx - cos80¢X*sinx/sinx
[(sin20¢X*sin10¢X/sin30¢X*sin40¢X) + cos80¢X]/sin80¢X= cotx
By calculator, x = 70¢X
------------------------------------- Finish ----------------------------------------
I want $$$ ar ~