trigonomertric formula

In triangle ABC, b=21mm, A=120 degrees and sinB : sinC=1 : 2
a) solve traingle ABC,
b) Let AD be the angle bisector of angleA (point D lies on BC), find the length of AD.

(a)
sinB : sinC=1 : 2
b/c=sinB/sinC=1/2
c=42 mm

BC^2=AB^2+AC^2-2(AB)(AC)cos120=3087
BC=55.5608 mm

And we can find angle B and C respectively, that is B=19.1066,C=40.8934

(b)

Using Angle bisector theorem

BD:BC=AB/AC=2
BD=(2/3)(√3087)
BC=(1/3)(√3087)

Using sine law

BD/sin60=AD/sinB
37.0405/0.8660=AD/0.3273
AD=14.0004 mm

2008-04-12 15:43:02 補充：
http://en.wikipedia.org/wiki/Angle_bisector_theorem

原來(b)有更簡單的方法﹐不過這條道理好有用﹐所以你有空的話不妨學識這招。

a)b / sinB = c / sinC (sine law)
21 / sinB = c / sinC
21 / sinB = c / sinC
21 / 1 = c / 2
c = 42 mm
a = √[b^2 + c^2 – 2bc cos A] (cosine law)
a = √[(21)^2 + (42)^2 – 2(21)(42) cos 120]
a = 55.6 mm
a / sinA = b / sinB
55.6 / sin120 = 21 / sinB
sinB = 0.3271
B = 19.1
C = 180 – 120 – 19.1
C = 40.9

2008-04-12 16:15:25 補充：
b)Since AD is the angle bisector of angleA
∠DAB = 60
∠ADB = 180–B–∠DAB
= 100.89
42/sinADB = AD/sinB
AD = 14mm