F4 A Maths

畫三角形計出cosA,cosB的值,這畫唔到圖,留給你畫
你會計到cosA=12/13 ,cosB=4/5
利用複角:
sinC=sin(A+B)
=sinAcosB+cosAsinB
=(5/13)(4/5)+(12/13)(3/5)
=56/65
b part 是可用計算機吧
三角形週界: AB+BC+AC=12 (1)

之後用SINE RULE
AC=ABsinB/sinC (2) BC=ABsinA/sinC (3)
SUB(2),(3)into (1)
AB+ABsinA/sinC+ABsinB/sinC =12
AB=5.6cm
方法應是咁吧

2007-04-15 16:12:47 補充：
因c隻角最大,佢對應的邊AB都係最大A,B都係銳角,加埋細過90,so 角c最大

a.sinC=sin[pi-(a+b)]=sin(A+B)//
sinA=5/13
so y(a)=5,r(a)=13
sinB=3/5
so y(b)=3,r(b)=5
sinC
=sin(A+B)
=sinA cos B+cosA sinB
=(5/13)(sqrt(r(b)^2-y(b)^2)/r(b))+(3/5)(sqrt(r(a)^2-y(a)^2)/r(a))
=(5/13)(4/5)+(3/5)(12/13)
=(20/65)+(36/65)
=56/65//
b. a:b:c=sinA:sinB:sinC
=5/13:3/5:56/65
=25:39:56
so c is the longest side
c=12x56/120
= 5.6//

sinC=sin[180 - (A+B)]
sinC = sin180cos(A+B) - cos180sin(A+B)
sinC = sin(A+B) since sin180=0 and cos180=-1

2007-04-15 15:58:12 補充：
sin(A+B) = sinAcosB + cosAsinBBy drawing right angled triangle, when sinA=5/13; cosA=12/13; sinB=3/5; cosB=4/5sinC=sin(A+B) =56/65

2007-04-15 16:00:29 補充：
let a/sinA = b/sinB = c/sinC = ka=(5/13)k ; b=(3/5)k ; c=(56/65)ka+b+c=12(5/13)k + (3/5)k + (56/65)k = 12

2007-04-15 16:07:32 補充：
[(5/13) + (3/5) + (56/65)]k = 12(120/65)k=12k=65/10a=(5/13)(65/10)=2.5b=(3/5)(65/10)=3.9c=(56/65)(65/10)=5.62.5+3.9+5.6=12c is the longest