Geometry with Trigonometry

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

= 1.