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From this result we can deduce two other results, in the same manner as (4) and (5) were deduced from (3); or we may observe that the right-hand member of (6) can be obtained from the right-hand member of (3) by writing 7-a and 1-6 for a and b respectively, and thus we may deduce the results more easily. We shall have then sino (1 C-E) =

cos , s sin 1 (s a) sin } (8 6) cos } (sc)

sin 1 a sin } o cosa coso (1 C-1 E)

sin 1 s cos 1 (s a) cos 1"(8 6) sin 1 (8 c)

sin } a sin 1 b cosc

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EXAMPLES.

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1. Find the angles and sides of an equilateral triangle whose area is one-fourth of that of the sphere on which it is described.

2. Find the surface of an equilateral and equiangular spherical polygon of n sides, and determine the value of each of the angles when the surface equals half the surface of the sphere.

7 3. If a=6=3, and ca

shew that E=cos 2'

9 4. If the angle C of a spherical triangle be a right angle, shew that sin 1 E=sin } a sin } b sec 1 c, cos 1 E=cos a cos } b sec } c. 5. If the angle C be a right angle, shew that

sino

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sinc

sino a

cos E =

COS C

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6. If a= b and C

shew that tan E

sino a 2 cos a

2'

7. The sum of the angles in a right-angled triangle is less than four right angles.

8. Draw through a given point in the side of a spherical triangle an arc of a great circle cutting off a given part of the triangle.

a

9. In a spherical triangle if cos C=-tan tanı, then

9

2

2

C= A + B.

10. If the angles of a spherical triangle be together equal to four right angles

cos* 1 a + cos* 1 b + cos' }c=1. 11. If r,, rg, rbe the radii of three small circles of a sphere of radius r which touch one another at P, Q, R, and A, B, C be the angles of the spherical triangle formed by joining their centres,

area PQR=(A cos r, + B cos r, +C cos rs – 7) **.

3

=

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12. Shew that

{ sin } E sin (A – } E) sin (B– } E) sin (C – } £}}}

E

sin s =

2 sin A sin B sin C

1

13. Given two sides of a spherical triangle, determine when the area is a maximum.

14. Find the area of a regular polygon of a given number of sides formed by arcs of great circles on the surface of a sphere; and hence deduce that, if a be the angular radius of a small circle, its area is to that of the whole surface of the sphere as versin a is to 2.

cos B'C

15. A, B, C are the angular points of a spherical triangle; A', B', C' are the middle points of the respectively opposite sides. If E be the spherical excess of the triangle, show that cos A'B'

cos C'A'
cos 1 E=
cos C cos a

cos 16. If one of the arcs of great circles which join the middle points of the sides of a spherical triangle be a quadrant, shew that the other two are also quadrants.

IX. ON CERTAIN APPROXIMATE FORMULÆ.

We shall now investigate certain approximate formulæ which are often useful in calculating spherical triangles when the radius of the sphere is large compared with the lengths of the sides of the triangles.

105. Given two sides and the included angle of a spherical triangle, to find the angle between the chords of these sides.

D

E

с

B

=

Let AB, AC be the two sides of the triangle ABC; let O be the centre of the sphere. Describe a sphere round A as a centre, and suppose it to meet A0, AB, AC at D, E, F respectively. Then the angle EDF is the inclination of the planes OAB, AC, and is therefore equal to A. From the spherical triangle DEF

cos EF = cos DE cos DF + sin DE sin DF cos A; and

DE=1 (T-c), DF= 1(7-6); therefore cos EF = sin 1 b sin c+cos 1 b cos c cos A.

If the sides of the triangle are small compared with the radius of the sphere, EF will not differ much from A; suppose EF=A-, then approximately

cos EF = cos A. +0 sin A ;

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O sin A = (1 - cos A) sino 1 (0+ c) – (1 + cos A) sino 4 (6-c), therefore 0 = tan 1 A sin 1 (6+c) – cot 1 A sin 1 (b-c).

a

This gives the circular measure of 0; the number of seconds in the angle is found by dividing the circular measure by the circular measure of one second, or approximately by the sine of one second (Plane Trigonometry, Art. 123). If the lengths of the arcs corresponding to a and b respectively be a and B, and r the radius of the

ß sphere, we have and as the circular measures of a and 6 respectively; and the lengths of the sides of the chordal triangle

B are 2r sin

and 2r sin respectively. Thus when the sides of

2r the spherical triangle and the radius of the sphere are known, we can calculate the angles and sides of the chordal triangle.

r

r

a

2r

106. Legendre's Theorem. If the sides of a spherical triangle be small compared with the radius of the sphere, then each angle of the spherical triangle exceeds by one third of the spherical excess the corresponding anyle of the plane triangle, the sides of which are of the same length as the arcs of the spherical triangle.

Let A, B, C be the angles of the spherical triangle ; a, b, c the sides; r the radius of the sphere; a, b, y the lengths of the arcs which form the sides, so that

are the circular

r measures of a, b, c respectively. Then

a

β γ

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Similar expressions hold for cos 6 and sin b, and for cos c and sin c respectively. Hence, if we neglect powers of the circular measure above the fourth, we have a a

B B4 1

ge 2474

27 2404 2702 24r* By

g2 1

1

6r2

a

-)(1

+

+

+ 272

COS A

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1
2By
ye - +
?

6r2
B + y -a

? a* + ß* +* – 2a*ß* 2B'ye – 2y*a? 2By

24Bype Now let A', B', C” be the angles of the plane triangle whose sides are a, B, y respectively; then

BP + ge - a

2By

9

COS A':

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Suppose A= A' + 0; then

cos A = cos A' - O sin A' approximately; therefore

By sin A'

s 0 = 6r?

Зr? ?

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