一條數學問題~~一條數學題

y+1=y/tan30'
Sol
tan30度=1/sqrt(3)
y+1=y/(1/sqrt(3))
y+1=sqrt(3)y
sqrt(3)y－y=1
y(sqrt(3)－1)=1
y=1/[sqrt(3)－1]
y=[sqrt(3)+1/[(sqrt(3)+1))sqrt(3)－1)]
y=(sqrt(3)+1)/2

y+1=y/tan30

since tan 30 = 1/√3

so the equation can be rewrite as the following :

y+1=y/(1/√3)

=>y+1=√3 y

=>(1-√3)y=-1

=>y=1/(√3-1)=(√3+1)/(3-1)=(√3+1)/2//

.........y + 1 = y/tan 30
.........y + 1 = y x 3^0.5
......1 + 1/y = 3^0.5
............1/y = (3^0.5) - 1
...............y = 1.366025~~

y+1=y/tan30'

tan30=1/sqrt(3)
y+1=y/tan30
=> y+1=y/(1/sqrt(3))
=> y+1=sqrt(3)y
=>1=[sqrt(3)-1]y
=>y=1/[sqrt(3)-1]
=>y=[sqrt(3)+1]/[sqrt(3)-1][sqrt(3)+1]
=>y=[sqrt(3)+1]/(3-1)
=>y=[sqrt(3)+1]/2 or 1.366

2009-06-20 21:55:15 補充：
y+1=y/tan30
y+1=y/0.5774 (你用計算機就知道tan30=0.5774)
y+1=1.732y
y+1-y=1.732y-y
1=0.732y
1/0.732=0.732y/0.732
1.366=y
y=1.366