Geometry with Trigonometry

42 Distance; degree-measure of an angle Ch. 3

A

B

C

H 1

k

Figure 3.7. Laying off an angle.

 A

B

C

B 1

C 1

Figure 3.8. Opposite angles at a vertex.

Degree-measure has the properties:-

(i)The null-angle∠BAB has degree-measure0.

(ii)For any non-null wedge-angle∠BAC, we have 0 <|∠BAC|◦<180.

(iii)If∠B 1 AC 1 is the angle vertically opposite to∠BAC, then

|∠B 1 AC 1 |◦=|∠BAC|◦,

so that vertically opposite angles have equal degree-measures.

Proof.

(i) LetCbe a point not onAB.ThenbyA 5 (iii) withD=B,

|∠BAB|◦+|∠BAC|◦=|∠BAC|◦.

It follows that|∠BAB|◦=0.
(ii) Given any non-null wedge-angle∠BAC,letH 1 be the closed half-plane with
edgeABin whichClies. If we had|∠BAC|◦=0, then we would have|∠BAC|◦=
|∠BAB|◦andsobyA 5 (iv) we would have[A,C=[A,B. This would imply that∠BAC
is null, contrary to assumption. Then by A 5 (i)|∠BAC|◦>0.
Now choose the pointE=Aso thatA∈[B,E].ThenbyA 5 (iii), as we have sup-
plementary angles,
|∠BAC|◦+|∠CAE|◦= 180.

But[A,E =[A,C as∠BACis a wedge-angle, so∠CAEis not a null-angle. By the
last paragraph we then have|∠CAE|◦>0 and it follows that|∠BAC|◦<180.
(iii) As|BAB 1 ,|CAC 1 are straight we have

|∠BAC|◦+|∠CAB 1 |◦= 180 ,

|∠CAB 1 |◦+|∠B 1 AC 1 |◦= 180 ,

there being two pairs of supplementary angles. It follows that

|∠BAC|◦+|∠CAB 1 |◦=|∠CAB 1 |◦+|∠B 1 AC 1 |◦,

from which we conclude by subtraction that|∠BAC|◦=|∠B 1 AC 1 |◦.