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therefore

=

a sin =hk - }(ko + k®) cos 0, nearly,
2hk - (k+k) (cos' 10-sinja)

2 sin

and

=

= 1 (h+k) tan 1 0–1 (h - k) cot 10. This process, by which we find the angle COD from the angle AOB, is called reducing an angle to the horizori.

XI. ON SMALL VARIATIONS IN THE PARTS OF A

SPHERICAL TRIANGLE.

128. It is sometimes important to know what amount of error will be introduced into one of the calculated parts of a triangle by reason of amy small error which may exist in the given parts. We will here consider an example,

129. A side and the opposite angle of a spherical triangle remain constant: determine the connexion between the small variations of any other pair of elements,

Suppose C and c to remain constant.
(1)

. Required the connexion between the small variations of the other sides. We suppose a and b to denote the sides of one triangle which can be formed with C and c as fixed elements, and a + da and 6 + 86 to đenote the sides of another such triangle ; then we require the ratio of da to sb. when both are extremely small. We have

a

and

COS C = cos a cos b + sin a sin b cos. C,
eos c = cos (a +8a) cos (6 +86) + sin (a + da) sin (6 +86) cos C;

cos (a + da) = cos a sin a da, nearly,
sin (a + da) = sin a + cos a da, nearly,

also

=

and

with similar formulæ for cos (b +86) and sin (b+86). (See Plane
Trigonometry, Chap. x11.) Thus
cos c= (cos a -- sin a da) (cos b - sin 6 86)

+(sin a + cos a da) (sin b + cos b 86) cos C.

6 Hence by subtraction, if we neglect the product da 86, 0 = da (sin a cos 6 – cos a sin b cos C)

+86 (sin b cos a cos b sin a cos C); this gives the ratio of da to 86 in terms of a, b, C. express the ratio more simply in terms of A and B; for, dividing by sin a sin b, we get from Art. 44,

We may

da

86
cot B sin C + cot A sin C= 0;
sin a

sin

therefore

da cos B + 86 cos A = 0.

(2) Required the connexion between the small variations of the other angles. In this case we may by means of the polar triangle deduce from the result just found, that

SA cos b + SB cos a =

0;

this may also be found independently as before.

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(3) Required the connexion between the small variations of

) a side and the opposite angle (A, a). Here

sin A sin c=sin C sin a, and

sin (A + 8A) sin c =sin C sin (a + da); hence by subtraction

cos A sin c A = sin C cos a da,

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(4) Required the connexion between the small variations of a side and the adjacent angle (a, b).

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We have cot C sin B = cot c sin a cos B cos a ; proceeding as before we obtain

cot C cos B&B = cot c cos a da + cos B sin a da + cos a sin B8B; therefore

(cot C cos B - cos a sin B) 8B = (cot c cos a + cos B sin a) da ;

a

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130. Some more examples are proposed for solution at the end of this Chapter; as they involve no difficulty they are left for the exercise of the student.

EXAMPLES.

86

+

2

sin c

1. In a spherical triangle, if C and c remain constant while a and b receive the small increments da and 86 respectively, shew that da

sin c

0 where n= J(1 nsinʼa) 71 - no sin 6) 2. If C and c remain constant, and a small change be made in a, find the consequent changes in the other parts of the triangle. Find also the change in the area.

3. Supposing A and c to remain constant, prove the following equations, connecting the small variations of pairs of the other elements :

sin C 86 = sin a SB, 86 sin C =- 8C tan a, da tan C=SB sin a, da tan C=-8C tan a, 86 cos C = da, SB cos a =- -8C.

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с

4. Supposing b and c to remain constant, prove the following equations connecting the small variations of pairs of the other elements : SB tan C = 8C tan B,

da cot C da = SA sin c sin B,

SA sin B cos C=-&B sin A.

8B sin a,

5. Supposing B and C to remain constant, prove the following equations connecting the small variations of pairs of the other elements : 86 tan c = dc tan b,

SA cot c = 86 sin A, 8A = da sin b sin C,

da sin B cos c =

86 sin A.

6. If A and C are constant, and 6 be increased by a small quantity, shew that a will be increased or diminished according as c is less or greater than a quadrant.

a

XII. ON THE CONNEXION OF FORMULÆ IN

PLANE AND SPHERICAL TRIGONOMETRY.

131. The student must have perceived that many of the results obtained in Spherical Trigonometry resemble others with which he is familiar in Plane Trigonometry. We shall now pay some attention to this resemblance. We shall first shew how we may deduce formulæ in Plane Trigonometry from formulæ in Spherical Trigonometry; and we shall then investigate some theorems in Spherical Trigonometry which are interesting principally on account of their connexion with known results in Plane Geometry and Trigonometry,

132. From any formula in Spherical Trigonometry involving the elements of a triangle, one of them being a side, it is required to deduce the corresponding formula in Plane Trigonometry.

Let a, B, y be the lengths of the sides of the triangle, r the radius of the sphere, so that

are the circular measures

α β γ

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of the sides of the triangle; expand the functions of

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a

β γ

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which occur in any proposed formula in powers of respectively; then if we suppose r to become indefinitely great,

1

the limiting form of the proposed formula will be a relation in Plane Trigonometry.

For example, in Art. 106, from the formula

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+

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COS A

pe+ge-a a* + Be +* - 2a®ß - 2By – 2y*a*

+ ; 2By

24Bypal now suppose r to become infinite; then ultimately

BP+ge - as

;

2By and this is the expression for the cosine of the angle of a plane triangle in terms of the sides. Again, in Art. 110, from the formula

sin A

6

sin a

sin B

sin

a

+

sin A a (ß-a") we deduce

+ ; sin BB

6Br? now suppose r to become infinite; then ultimately

sin A

sin BB ' that is, in a plane triangle the sides are as the sines of the opposite angles.

a

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133. To find the equation to a small circle of the sphere. The student can easily draw the required diagram.

Let O be the pole of a small circle, S a fixed point on the sphere, SX a fixed great circle of the sphere. Let OS = a, OSX = p; then the position of O is determined by means of these angular co-ordinates a and ß. Let P be any point on the circumference of the small circle, PS = 0, PSX = $, so that @ and are T. S. T.

H

a

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