解三角學的問題,~急~

圖片參考：http://i320.photobucket.com/albums/nn347/old-master/20091120_1.jpg?t=1258701191

http://i320.photobucket.com/albums/nn347/old-master/20091120_1.jpg?t=1258701191

2009-11-20 15:57:22 補充：
另一方法是作AH垂直於AC，垂足為H。

ΔBCH中：
∠CBH = 5.5º
BH = BCcos5.5º
HC = BCsin5.5o

ΔABH中：
AH = BH/tan30º = BCcos5.5º/tan30º

AH + HC = 109 m
BCcos5.5º/tan30º + BCsin5.5o = 109 m
BC = 109/[(cos5.5º/tan30º) + sin5.5o] m = 60 m
不傾斜時，塔高 = 60 m

109tan(30度)sec(5.5度)
=63(m)

2009-11-20 14:14:30 補充：
另一解

設塔高x
[xsin(5.5 度)+109]tan(30度)=xcos(5.5度)
109tan(30度)=x[cos(5.5度)-sin(5.5度)]
x=69.95約=70(m)

2009-11-20 14:17:33 補充：
另一解
[xsin(5.5度)+109]tan(30度)=xcos(5.5度)
109tan(30度)=x[cos(5.5度)- sin(5.5度)]
x=69.95約=70(m)

2009-11-20 14:17:50 補充：
另一解
[xsin(5.5度)+109]tan(30度)=xcos(5.5度)
109tan(30度)=x[cos(5.5度)- sin(5.5度)]
x=69.95約=70(m)