Geometry with Trigonometry

170 Complex coordinates; sensed angles; angles between lines Ch. 10

We extend our frame of referenceFby taking in connection with the line pair
{OI,OJ}a canonical pair of duo-sectorsD 1 andD 2 , withD 1 the union of the first
and third quadrantsQ 1 andQ 3 ,andD 2 the union of the second and fourth quadrants.

For any linelthrough the originO, we consider the duo-angleαdwith side-lines
OIandl, such that the indicatormofαdlies in the duo-sectorD 1 , that is the bisector
of the line-pair{OI,l}which lies in the duo-sector ofαdalso lies inD 1. We denote
byDA(F)the set of such duo-angles, and we say that they are instandard position
respect toF.
Ifl=OIandZ 4 ≡(x 4 ,y 4 )is a point other thanOonl, then so is the point with
coordinates(−x 4 ,−y 4 ); thus, without loss of generality, we may assume thaty 4 > 0
in identifyinglasOZ 4 .ThenZ 4 ∈H 1 and

|αd|◦=|∠IOZ 4 |◦,

cosαd=cos(∠IOZ 4 )=

x 4

√

x^24 +y^24

,

sinαd=sin(∠IOZ 4 )=

y 4

√

x^24 +y^24

.

Whenαdis not a right duo-angle, we have

tanαd=

y 4

x 4

.

We identifyl=OIasOZ 4 whereZ 4 ≡(x 4 , 0 )andx 4 >0. Thus for the null duo-
angle in standard position we have cosαd= 1 ,sinαd= 0 ,tanαd=0. We denote this
null duo-angle by 0dFand the right duo-angle in standard position by 90dF.

We now note that ifαd,βd∈DA(F)andtanαd=tanβd,thenαd=βd.
Proof. For this we letαd,βdhave pairs of side-lines(OI,OZ 4 ),(OI,OZ 5 ), respec-
tively, where|O,Z 4 |=|O,Z 5 |=k, and eithery 4 >0orx 4 > 0 ,y 4 =0, and similarly
eithery 5 >0orx 5 > 0 ,y 5 =0. Then neitherαdnorβdis 90dFand

y 4

x 4

=

y 5

x 5

,x^24 +y^24 =x^25 +y^25 =k^2.

Ify 4 =0theny 5 =0 and both duo-angles are null. Suppose then thaty 4 =0so
thaty 4 >0; it follows thaty 5 >0. Then

k^2 =x^25 +y^25 =

y^25

y^24

x^24 +y^25 =

y^25

y^24

(x^24 +y^24 )=

y^25

y^24

k^2.

Hencey^25 =y^24 ,andsoy 5 =y 4. It follows thatx 5 =x 4.