Sec. 10.8 Sensed angles of triangles, the sine rule 163
For non-collinear points Z 1 ,Z 2 ,Z 3 if
α=FZ 2 Z 1 Z 3 ,β=FZ 3 Z 2 Z 1 ,γ=FZ 1 Z 3 Z 2 ,
andθ=αF,φ=βF,ψ=γF,thenθ+φ+ψ= (^180) F.
Proof.For
z 3 −z 1
z 2 −z 1
=
b
c
cisθ,
z 1 −z 2
z 3 −z 2
=
c
a
cisφ,
z 2 −z 3
z 1 −z 3
=
a
b
cisψ.
On multiplying these together, we find that
− 1 =cisθ.cisφ.cisψ=cis(θ+φ+ψ).
As cis180F=−1 it follows thatθ+φ+ψ= (^180) F.
With the above notation, the lengths of the sides and the sensed-angles of a trian-
gle[Z 1 ,Z 2 ,Z 3 ]have the properties:-
(i)In each case
vcisβ=
1
1 −ucisα
,
and two pairs of similar identities obtained from these on advancing cyclically
through(u,v,w)and(α,β,γ).
(ii)In each case
c=bcosα+acosβ,
sinα
a
=
sinβ
b
,
and two pairs of similar identities obtained from these on advancing cyclically
through(a,b,c)and(α,β,γ).
Proof.
(i) Forz 3 −z 1 =ucisα.(z 2 −z 1 )so thatz 3 −z 2 =( 1 −ucisα)(z 1 −z 2 ), while
z 1 −z 2 =vcisβ.(z 3 −z 2 ),which give( 1 −ucisα)vcisβ= 1.
(ii) From (i)
1 −u[cosα+ısinα]=
1
v
[cosβ−ısinβ],
so equating real parts givesc=bcosα+acosβ,while equating imaginary parts
gives sinα/a=sinβ/b.
This result re-derives the sine rule for a triangle.
If Z 1 ,Z 2 ,Z 3 are distinct points, then
Z 3 Z (^1) F
Z 3 Z 2 F
+Z^2 Z^1 F
Z 2 Z 3 F
= 1.
Proof.For
z 1 −z 3
z 2 −z 3
+
z 1 −z 2
z 3 −z 2