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SPHERICAL TRIGONOMETRY.

I. It has already been shown that a spherical triangle is formed by the arcs of three great circles intersecting each other on the surface of a sphere, (Book IX. Def. 1). Hence, every spherical triangle has six parts: the sides and three angles.

Spherical Trigonometry explains the methods of determining, by calculation, the unknown sides and angles of a spherical triangle when any three of the six parts are given.

II. Any two parts of a spherical triangle are said to be of the same species when they are both less or both greater than 90°; and they are of different species when one is less and the other greater than 90°.

III. Let ABC be a spherical triangle, and O the centre of the sphere. Let the sides of the triangle be designated by letters corresponding to their opposite angles: that is, the side opposite B the angle A by a, the side opposite B by b, and the side opposite C by c. Then the angle COB will be represented by a, the angle COA by b and the angle BOA by c. The angles of the

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H

D

F

a

spherical triangle will be equal to the angles included between the planes which determine its sides (Book IX. Prop. VI.).

From any point A, of the edge OA, draw AD perpendicular to the plane COB. From D draw DH perpendicular to OB, and DK perpendicular to OC; and draw AH and AK: the last lines will be respectively perpendicular to OB and OC, (Book VI. Prop. VI.)

The angle DHA will be equal to the angle B of the spherical triangle, and the angle DKA to the angle C.

The two right angled triangles OKA, ADK, will give the proportions

R: sin AOK :: OA: AK, or, Rx AK-OA sin b.

R: sin AKD :: AK: AD, or, R× AD=AK sin C.

Hence, R2 × AD=AO sin b sin C, by substituting for AK its value taken from the first equation.

In like manner the triangles AHO, ADH, right angled at H and D, give

R: sin c:: AO: AH, or Rx AH-AO sin c
R: sin B:: AH: AD, or RX AD AH sin B.
Hence, R2× AD=AO sin c sin B.

Equating this with the value of R2 viding by AO, we have

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AD, before found, and di

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sin B sin C:: sin b: sin c that is,

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The sines of the angles of a spherical triangle are to each other as the sines of their opposite sides.

IV. From K draw KE perpendicular to OB, and from D draw DF parallel to OB. Then will the angle DKF-COB=a, since each is the complement of the angle EKO. In the right angled triangle OAH, we have

R: cos c: OA: OH; hence

AO cos c=RX OH=RXOE+R.DF.

In the right-angled triangle OKE

R: cos a: OK : OE, or R× OE=OK cos a.

But in the right angled triangle OKA

R: cos b:: OA: OK, or, Rx OK-OA cos b.

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In the right-angled triangle KFD

R sin a KD: DF, or Rx DF-KD sin a.
But in the right angled triangles OAK, ADK, we have
R sin b:: OA : AK, or R× AK=OA sin b
R: cos K: AK: KD, or R× KD=AK cos C

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Similar equations may be deduced for each of the other sides. Hence, generally,

R2 cos a=R cos b cos c+sin b sin c cos A.
R2 cos b-R cos a cos c+ sin a sin c cos B.
R2 cos c=R cos b cos a+sin b sin a cos C.

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(2.)

That is, radius square into the cosine of either side of a spherical triangle is equal to radius into the rectangle of the cosines of the two other sides plus the rectangle of the sines of those sides into the cosine of their included angle.

V. Each of the formulas designated (2) involves the three sides of the triangle together with one of the angles. These formulas are used to determine the angles when the three sides are known. It is necessary, however, to put them under another form to adapt them to logarithmic computation.

Taking the first equation, we have

cos A

R2 cos a-R cos b cos c
sin b sin c

Adding R to each member, we have

R2 cos a+R sin b sin c-R cos b cos c
sin b sin c

R+cos A=

2 cos A

But, R+cos A

R

(Art. XXIII.), and

R sin b sin c-R cos b cos c=-] -R2 cos (b+c) (Art. XIX.);

hence,

2 R

2 cos2A_R2 (cos a-cos (b+c)).

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Putting s=a+b+c, we shall have

(Art. XXIII).

s=(a+b+c) and 1s—a= (b+c—a): hence

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Had we subtracted each member of the first equation from R, instead of adding, we should, by making similar reductions, have found

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sin a sin c

sin } B=R√sin}(a+b—c) sin }; (b+c—a)

sinC=RV

sin(a+c-b) sin (b+c-a)

sin a sin b

Putting s=a+b+c, we shall have

(4.)

}s—a={(b+c—a), }s—b= (a+c-b), and s-c=(a+b—c)

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VI. We may deduce the value of the side of a triangle in terms of the three angles by applying equations (4.), to the polar triangle. Thus, if a, b, c, A', B', C', represent the sides and angles of the polar triangle, we shall have

A=180°-a', B=180°-b', C=180°--c';

a=180°-A', b=180°-B', and c=180°-C'

(Book IX. Prop. VII.): hence, omitting the ', since the equations are applicable to any triangle, we shall have

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cos (A+B-C) cos (A+C-B)
sin B sin C

cos (A+B—C) cos (B+C-A) (6.)

sin A sin C

cos c=Rcos (A+C-B) cos (B+C—A)

sin A sin B.

Putting S=A+B+C, we shall have

S—A=}(C+B—A), }S—B=} (A+C—B) and }S—C=}(A+B-C), hence

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VII. If we apply equations (2.) to the polar triangle, we shall have

—R2 cos A'=R cos B' cos C'-sin B' sin C' cos a'.

Or, omitting the', since the equation is applicable to any triangle, we have the three symmetrical equations,

R2.cos A=sin B sin C cos a- -R cos B cos C

R2.cos B sin A sin C cos b-R cos A cos C (8.)

B-sin

R2.cos C sin A sin B cos c—]
-R cos A cos B

That is, radius square into the cosine of either angle of a spherical triangle, is equal to the rectangle of the sines of the two other angles into the cosine of their included side, minus radius into the rectangle of their cosines.

VIII. All the formulas necessary for the solution of spherical triangles, may be deduced from equations marked (2.). If we substitute for cos b in the third equation, its value taken from the second, and substitute for cos a its value R2-sin2 a, and then divide by the common factor R.sin a, we shall have R.cos c sin a sin c cos a cos B+R.sin b cos C.

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