f.3 maths 急

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假設 AB 長度為 r cm。
那麼 AC = AB = r cm。

從 A 拉一條直線到 BC 與 BC 垂直，交點為 D。因為 AD 與 BC 垂直，所以 AD 是 △ABC 的高度。

同時 cos B = BD/AB

cos 30° = BD/r
√3/2 = BD/r
BD = √3r/2

因為 AD 垂直 BC，所以角 ADB = 角 ADC = 90°，
同時 AB = AC，
同時 AD 時同邊，
所以 △ABD 全等 △ACD。

所以 BD = CD = √3r/2 cm。
BC
= BD + CD
= √3r/2 + √3r/2
= √3r cm

AB + BC + CA = 10 cm
r + √3r + r = 10
(2+√3)r = 10
r = 10/(2+√3)

所以
AB
= r
= 10/(2+√3)
= 2.679 cm

AC
= AB
= 10/(2+√3)
= 2.679 cm

BC
= √3(10/(2+√3))
= 4.641 cm

AB = AC = 2.6795 cm
BC = 4.6410 cm

use formula a^2=b^2+c^2- 2bc(cosA) and b=c since B=C=30degree triangle=180degree therefore A=120degree
therefore a^2=2b^2 - 2b^2 cos120
a^2=2b^2 - 2b^2(-0.5)
a^2=2b^2-(-b^2)
a^2=3b^2
a =3^0.5(b)
A+B+C=10 therefore a+b+c=10__ c=b __ a+2b=10__a=10-2b
3^0.5b=10-2b 3^0.5=1.732
1.732b+2b=10
b(1.732+2)=10
b=2.68
2.68=b=c or 2.68=AB=AC
a=10-2b
a =10-2(2.68)
a=4.64

∠C = ∠B = 30∘ (base ∠s, isos.Δ)
In ΔABC, Draw AD ⊥ BC

In rt.Δ ADB
BD/ AB = cos∠B
BD/ AB = cos30∘
BD/ AB = √3/2
BD = (√3/2) AB

In ΔABC,
AC = AB (given)
BD = DC (property if isos.Δ)

Consider the perimeter of ΔABC,
AC + AB + BD + DC = 10
AB + AB + BD + BD = 10
2AB + 2BD = 10
2AB + 2x (√3/2) AB = 10
2AB +√3 AB = 10
AB (2+√3 ) = 10
AB = 10/(2+√3 )
AB = 2.6795 cm

AC = 2.6795 cm

BC
= 10 - AB - AC
= 10 - 2 x 2.6795
= 4.6410 cm


ANS:

AB = AC = 2.6795 cm
BC = 4.6410 cm

１＞AC = AB ( isosceles triangle )
２＞角ABC = 角ACB = 30度 （isosceles triangle）
３＞角BAC = 180 度 - 30 度 - 30 度 ( sum of triangle )
= 120度

－－－－－－－－－－－－－－－－－－－－－－－－

let AC be x cm,
AB = x
BC = 10 - 2x

　ｘ　　　　10-2x
－－－　＝　－－－
sin 30度　　sin 120度

x sin 120度 = (10 - 2x) sin 30度
sin 120度 = 0.5(10) - 0.5(2x)
sin 120度 x = 5 - x
(1 + sin 120度) x = 5
x = 2.68 cm

BC = 10 - 2x = 4.64cm
AB = AC = 2.68cm

2006-11-27 00:37:30 補充：
y99shto has done a good job!