Geometry with Trigonometry

Sec. 10.4 Sensed angles 155

(ii) From (i)

ℑ

z 2 −z 0

z 1 −z 0

=

|z 2 −z 0 |

|z 1 −z 0 |

sin(θ 2 −θ 1 ),

and this is positive or negative according asθ 2 −θ 1 is wedge or reflex. Moreover

|z 1 −z 0 |^2 ℑ

z 2 −z 0

z 1 −z 0

=ℑ[(z 2 −z 0 )(z ̄ 1 −z ̄ 0 )] = (y 2 −y 0 )(x 1 −x 0 )−(x 2 −x 0 )(y 1 −y 0 ).

(iii) For the first isθ 2 −θ 1 and the second isθ 1 −θ 2.

(iv) For ifθ 1 =∠F′I 0 Z 0 Z 1 ,θ 2 =∠F′I 0 Z 0 Z 2 ,θ 3 =∠F′I 0 Z 0 Z 3 ,then

θ 2 −θ 1 +(θ 3 −θ 2 )=θ 3 −θ 1.

(v) For ifZ∈[Z 0 ,Z 1 ,thenz=z 0 +t(z 1 −z 0 )=z 0 +t|z 1 −z 0 |cisθ 1 for somet≥0.
Hencerθ 2 −θ 1 ;Z 0 (Z)=Z′where

z′−z 0 =(z−z 0 )cis(θ 2 −θ 1 )=t|z 1 −z 0 |cisθ 1 cis(θ 2 −θ 1 )=t|z 1 −z 0 |cisθ 2.

ThusZ′∈[Z 0 ,Z 2.
(vi) Forθ′−θ=α.
If the points Z 1 and Z 2 are both distinct from Z 0 andφ=FZ 1 Z 0 Z 2 ,then

|Z 1 ,Z 2 |^2 =|Z 0 ,Z 1 |^2 +|Z 0 ,Z 2 |^2 − 2 |Z 0 ,Z 1 ||Z 0 ,Z 2 |cosφ.

Proof. For by (i) in the last result,

z 2 −z 0 =

|Z 0 ,Z 2 |

|Z 0 ,Z 1 |

(cosφ+ısinφ)(z 1 −z 0 ),

so that

z 2 −z 1 =

[

|Z 0 ,Z 2 |

|Z 0 ,Z 1 |

(cosφ+ısinφ)− 1

]

(z 1 −z 0 ).

Then

|Z 1 ,Z 2 |^2 =

{[

|Z 0 ,Z 2 |

|Z 0 ,Z 1 |

cosφ− 1

] 2

+

[

|Z 0 ,Z 2 |

|Z 0 ,Z 1 |

sinφ

] 2 }

|Z 0 ,Z 1 |^2 ,

and the result follows on expanding the right-hand side here.

For a non-collinear triple(Z 0 ,Z 1 ,Z 2 ),letαbe the wedge-angle∠Z 1 Z 0 Z 2 andφ
be the sensed angleFZ 1 Z 0 Z 2 .Thencosφ=cosαso that|φ|◦=|α|◦whenφis
wedge, and|φ|◦= 360 −|α|◦whenφis reflex.
Proof. By the last result,

|Z 1 ,Z 2 |^2 =|Z 0 ,Z 1 |^2 +|Z 0 ,Z 2 |^2 − 2 |Z 0 ,Z 1 ||Z 0 ,Z 2 |cosφ,

while by the cosine rule for a triangle in 9.5.1

|Z 1 ,Z 2 |^2 =|Z 0 ,Z 1 |^2 +|Z 0 ,Z 2 |^2 − 2 |Z 0 ,Z 1 ||Z 0 ,Z 2 |cosα.

Hence cosφ=cosαso that sin^2 φ=sin^2 αand hence sinφ=±sinα. The result
follows from 9.3.4 and 9.6.