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the angular co-ordinates of P. Let OP=r. Then from the triangle OSP

cos r= cos a cos 6 + sin a sin 8 cos ($-B)............ (1); this gives a relation between the angular co-ordinates of any point on the circumference of the circle.

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If the circle be a great circle then r= ; thus the equation becomes

0 = cos a cos 0 + sin a sin 6 cos ($-B)......... (2). It will be observed that the angular co-ordinates here used are analogous to the latitude and longitude which serve to determine the positions of places on the Earth's surface ; & is the complement of the latitude and $ is the longitude.

134, Equation (1) of the preceding Article may be written

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Divide by cos', and

hence rearrange;

2 0

A tan 5 (cos r + cos a) – 2 tan sin a cos ($-B) + COS r – - cos a = 0.

2

9, and tan

0 Let tan and tan denote the values of tan found from 2

2 this quadratic equation; then by Algebra, Chapter XXII. , .

atr
tan tan

tan tan
2
ÇOS " + cos a

2

COST- cos a

a-1

2

.

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Thus the value of the product tan tan is independent of $;

2 2 this result corresponds to the well-known property of a circle in Plane Geometry which is demonstrated in Euclid III, 36 Corollary.

135. Let three arcs OA, OB, OC meet at a point. From any point P in OB draw PM perpendicular to 04, and PN perpendicular to OC. The student can easily draw the required diagram,

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Then, by Art. 65,
sin PM = sin OP sin AOB, sin PN=sin OP sin COB;

sin PM sin AOB therefore

sin PN

sin COB Thus the ratio of sin PM to sin PN is independent of the position of P on the arc OB.

136. Conversely suppose that from any other point p arcs pm and pn are drawn perpendicular to OA and OC respectively; then if

sin PM
sin PN'

sin pm sin pn

it will follow that p is on the same great circle as O and P.

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and a

sin x=

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sin de,

1

+

sin Xu

1

1

137. From two points P and P, arcs are drawn perpendicular to a fixed arc; and from a point P on the same great circle as P, and P, a perpendicular is drawn to the same fixed arc. Let PP, = 0, and PP, = 0,; and let the perpendiculars drawn from P, P,, and P, be denoted by 2, 2,

Then will sin .

sin , sin (0, +) sin (@, + ) Let the arc P,P,, produced if necessary, cut the fixed arc at a point 0; let a denote the angle between the arcs.

We will suppose that P, is between 0 and Ps, and that P is between P, and P, Then, by Art. 65, sin x, = sin a sin OP, = sin a sin (OP-0,)

=sin a (sin OP cos 0, - cos OP sin 0,); sin x, = sin a sin OP, = sin a sin (OP+0)

= sin a (sin OP cos 0 + cos OP sin 0,). Multiply the former by sin 0,, and the latter by sin 0,, and add ; thus sin 0, sin x, + sin 0, sin x, = sin (0, +0%) sin a sin OP = sin (@, +0%) sin ..

a

=

a

1

The student should convince himself by examination that the result holds for all relative positions of P, P, and P, when due regard is paid to algebraical signs,

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138. The principal use of Art. 137 is to determine whether three given points are on the same great circle; an illustration will be given in Art. 146.

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139. The arcs drawn from the angles of a spherical triangle perpendicular to the opposite sides respectively meet at a point.

E

F.

B

D

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sinn

cos B

Let CF be perpendicular to AB. From F suppose arcs drawn perpendicular to CB and CA respectively; denote the former by ¿ and the latter by n. Then, by Art. 135,

sin É sin FCB

sin FCA But, by Art. 65,

= cos CF sin FCB, = cos CF sin FCA;
sing

cos B cos C therefore

cos A cos C° And if from any point in CF arcs are drawn perpendicular to CB and CA respectively, the ratio of the sine of the former perpendicular to the sine of the latter perpendicular is equal to by Art. 135.

COS A:

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sin ๆ

sin & sinn

In like manner suppose AD perpendicular to BC; then if from any point in AD arcs are drawn perpendicular to AC and AB respectively, the ratio of the sine of the former perpendicular to the sine of the latter perpendicular is equal to

cos A cos C
cos A cos B

Let CF and AD meet at P, and from P let perpendiculars be drawn on the sides a, b, c of the triangle; and denote these perpendiculars by x, y, z respectively : then we have shewn that

cos B cos C sin ly cos A cos ?

sin a

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sin a

sin z

cos B cos C

cos B cos A' and this shews that the point P is on the arc drawn from B perpendicular to AC.

Thus the three perpendiculars meet at a point, and this point is determined by the relations

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140. In the same manner it may be shewn that the arcs drawn from the angles of a spherical triangle to the middle points of the opposite sides meet at a point; and if from this point arcs X, Y, are drawn perpendicular to the sides a, b, c respectively,

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141. It is known in Plane Geometry that a certain circle touches the inscribed and escribed circles of any triangle; this circle is called the Nine points circle : see Appendix to Euclid, pages 317, 318, and Plane Trigonometry, Chapter xxiv.

We shall now shew that a small circle can always be determined on the sphere to touch the inscribed and escribed circles of any spherical triangle.

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142. Let a denote the distance from A of the pole of the small circle inscribed within a spherical triangle ABC. Suppose that a small circle of angular radius p touches this inscribed circle

P internally; let ß be the distance from A of the pole of this touching circle ; let y be the angle between arcs drawn from A to the pole of the inscribed circle and the pole of the touching circle respectively. Then we must have

cos (p-1) = cos a cos ß + sin a sin ß cos y ...........(1). Suppose that this touching circle also touches externally the escribed circle of angular radius ni;

then if

a,

denote the distance from A of the pole of this escribed circle, we must have

cos (+9)= cos a, cos ß + sin a, sin ß cos y .... Similarly, if a, and a, denote the distances from A of the poles of the other escribed circles, in order that the touching circle may touch these escribed circles externally, we must also have

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T

cos (p +r.) = cos a, cos B + sin a sin B cos (1 –Y).

v)........ (3). cos (p+r.) = cos a, cos B + sin a, sin B cos (+ y )...... (4).

We shall shew that real values of p, B, and y can be found to satisfy these four equations. Eliminate cos y from (1) and (2); thus

y cosp (cos r sin' a, - cos r, sin a) + sin p (sin r sin a, + sin r, sin a) = cos ß (cos a sin a, - cos an

(5).

sin a)

Suppose that the inscribed circle touches AB at the distance m from A, and that the escribed circle of angular radius touches AB at the distance m, from A. Then, by Art. 65,

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