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(3) Suppose that one of the sides which contain the angle A is a quadrant, for example, AB; on AC, produced if necessary,

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a

take AD equal to a quadrant and draw BD. If BD is a quadrant B is a pole of AC (Art. 11); in this case a =

and A 2

2

T

T

as well

T as C=

2

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Thus the formula to be verified reduces to the identity 0 = 0. If BD be not a quadrant, the triangle BDC gives

cos a = cos CD cos BD + sin CD sin BD cos CDB, and cos CDB= 0, cos CD= cos b sin b, cos BD = cos A ;

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thus

cos a= sin b cos A ; and this is what the formula in Art. 37 becomes when c=

2

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(4) Suppose that both the sides which contain the angle A are quadrants. The formula then becomes cos arcos A ; and this is obviously true, for A is now the pole of BC, and thus A = a.

Thus the formula in Art. 37 is proved to be universally true.

39. The formula in Art. 37 may be applied to express the cosine of any angle of a triangle in terms of sines and cosines of the sides ; thus we have the three formulæ,

cos a = cos 6 cos c + sin b sin c cos A,

b = = cos c cos a + sin c sin a cos B, COS C = cos a cos 6+ sin a sin b cos C.

COS

a

These may be considered as the fundamental equations of Spherical Trigonometry; we shall proceed to deduce various formula from them.

• 40. To express the sine of an angle of a spherical triangle in terms of trigonometrical functions of the sides.

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therefore

с

sin A = /(1- cos’a – cos’6 – cos'c+2 cos a cosb cose)

sin b sin c

The radical on the right-hand side must be taken with the positive sign, because sin b, sin c, and sin A are all positive.

41. From the value of sin A in the preceding Article it follows that

sin A sin B sin c
sin a
sin 6

sin c

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for each of these is equal to the same expression, namely,

N(1 - cos'a – cos®6 cos®c + 2 cos a cos b cos c)

sin a sin b sin c

Thus the sines of the angles of a spherical triangle are proportional to the sines of the opposite sides. We will give an independent proof of this proposition in the following Article.

42. The sines of the angles of a spherical triangle are proportional to the sines of the opposite sides.

Let ABC be a spherical triangle, O the centre of the sphere. Take any point P in 04, draw PD perpendicular to the plane

B

F

S

BOC, and from D draw DE, DF perpendicular to OB, OC respectively; join PE, PF, OD.

Since PD is perpendicular to the plane BOC, it makes right angles with every straight line meeting it in that plane; hence

PE` = PDS + DE= POS - OD+ DE = POOE? ; thus PEO is a right angle. Therefore PE=OP sin POE=OP sin c; and PD=PE sin PED=PE sin B=OP sin c sin B. Similarly, PD'= OP sin b sin C; therefore

OP sin c sin B = OP sin b sin C;

sin B sin 6 therefore

sin C sin c. The figure supposes b, c, B, and C each less than a right angle;

C it will be found on examination that the proof will hold when the figure is' modified to meet any case which can occur.

If, for instance, B alone is greater than a right angle, the point D will fall beyond. OB instead of between OB and OC; then PED will be the supplement of B, and thus sin PED is still equal to sin B.

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43. To shew that cot a sin b = cot A sin C + cos b cos C. We have cos a = cos b cos C + sin b sin c cos A, COS C = cos a cos + sin a sin 6 cos C,

sin C sin c = sin a

sin A

Substitute the values of cos c and sin c in the first equation ; thus

sin a sin 6 cos A sin C cos a = (cos a cos 6 + sin a sin 6 cos C) cos b +

;

sin A by transposition

cos a sino = sin a sin b cos 6 cos C + sin a sin b cot A sin C ; divide by sin a sin b; thus

cot a sin 6 = cos 6 cos C + cot A sin C.

2

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44. By interchanging the letters five other formulæ may be obtained like that in the preceding Article; the whole six formulæ will be as follows:

cot a sin b = cot A sin C + cos 6 cos C,
cot b sin a= cot B sin C + cos a cos C,
cot b sin c cot B sin A + cos c cos A,
cot c sin b = cot C sin A + cos b cos A,
cot c sin a =

cot C sin B + cos a cos B,

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cot a sin c= cot A sin B + cos c cos B.

45. To express the sine, cosine, and tangent, of half an angle of a triangle as functions of the sides.

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Let 28= a + b +c, so that s is half the sum of the sides di the triangle; then a+b-c=28 - 2c = 2 (8-c), a-b+c= 28 – 26= 2 (8 - 0);

s A sin (8-6) sin (8 - c) thus

sin 2

sin b sinc

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The positive sign must be given to the radicals which occur in

А this Article, because is less than a right angle, and therefore its

2 sine, cosine, and tangent are all positive.

46. Since sin A = 2 sin

A А
COS

we obtain

2

sin A

sin bsin e {sin s sin (8– a) sin (8 – b) sin (8-0);?

c

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c

It may be shewn that the expression for sin A in Art. 40 agrees with the present expression by putting the numerator of that expression in factors, as in Plane Trigonometry, Art. 115. We shall find it convenient to use a symbol for the radical in the value of sin A ; we shall denote it by n, so that

ne = sin s sin (s – a) sin (8 6) sin (8- c),

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and

4n = 1- cos'a - cos 6 – cos'c + 2 cos a cos 6 cos c.

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