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For the sum of the three plane angles which form the solid angle at O is less than four right angles (Euclid, xi. 21); therefore AB BC CA

is less than 21,
ОА
ОА

ОА

+

+

therefore, AB + BC + CA is less than 27 ROA; that is, the sum of the arcs is less than the circumference of a great circle.

31. The propositions contained in the preceding two Articles may be extended. Thus, if there be any polygon which has each of its angles less than two right angles, any one side is less than the sum of all the others. This may be proved by repeated use of Art. 29. Suppose, for example, that the figure has four sides, and let the angular points be denoted by A, B, C, D. Then

AB + BC is greater than AC; therefore, AB + BC + CD is greater than AC + CD, and à fortiori greater than AD,

Again, if there be any polygon which has each of its angles less than two right angles, the sum of its sides will be less than the circumference of a great circle. This follows from Euclid, xi. 21, in the manner shewn in Art. 30.

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The three angles of a spherical triangle are together greater than two right angles and less than six right angles.

Let A, B, C be the angles of a spherical triangle; let a', b', c' be the sides of the polar triangle. Then by Art. 30,

á + b' + c' is less than 27, that is,

7 - A+B+-C is less than 27; therefore, A + B + C is greater than a. And since each of the angles A, B, C is less than Tg

the A + B + C is less than 37.

T

sum

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33. The angles at the base of an isosceles spherical triangle are equal.

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Let ABC be a spherical triangle having AC = BC; let O be the centre of the sphere. Draw tangents at the points A and B to the arcs AC and BC respectively; these will meet OC produced at the same point S, and AS will be equal to BS.

Draw tangents AT, BT at the points A, B to the arc AB; then AT=TB; join TS. In the two triangles SAT, SBT the sides SA, AT, TS are equal to SB, BT, T'S respectively; therefore the angle SAT is equal to the angle SBT'; and these are the angles at the base of the spherical triangle.

The figure supposes AC and BC to be less than quadrants ; if they are greater than quadrants the tangents to AC and BC will meet on CO produced through 0 instead of through C, and the demonstration may be completed as before. If AC and BC are quadrants, the angles at the base are right angles by Arts. 11 and 9,

34. If two angles of a spherical triangle are equal, the opposite sides are equal.

Since the primitive triangle has two equal angles, the polar triangle has two equal sides ; therefore in the polar triangle the angles opposite the equal sides are equal by Art. 33. Hence in the primitive triangle the sides opposite the equal angles are equal.

35. If one angle of a spherical triangle be greater than another, the side opposite the greater angle is greater than the side opposite the less angle.

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Let ABC be a spherical triangle, and let the angle ABC be greater than the angle BAC : then the side AC will be greater than the side BC. At B make the angle ABD equal to the angle BAD; then BD is equal to AD (Art. 34), and BD + DC is greater than BC (Art. 29); therefore AB+ DC is greater than BC; that is, AC is greater than BC.

36. If one side of a spherical triangle be greater than another, the angle opposite the greater side is greater than the angle opposite the less side.

This follows from the preceding Article by means of the polar triangle.

Or thus; suppose the side AC greater than the side BC, then the angle ABC will be greater than the angle BAC. For the angle ABC cannot be less than the angle BAC by Art. 35, and the angle ABC cannot be equal to the angle BAC by Art. 34; therefore the angle ABC must be greater than the angle BAC.

This Chapter might be extended ; but it is unnecessary to do so because the Trigonometrical formulæ of the next Chapter supply an easy method of investigating the theorems of Spherical Geometry. See Arts. 56, 57, and 58.

IV. RELATIONS BETWEEN THE TRIGONOMETRICAL

FUNCTIONS OF THE SIDES AND THE ANGLES
OF A SPHERICAL TRIANGLE.

37. To express the cosine of an angle of a triangle in terms of sines and cosines of the sides.

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Let ABC be a spherical triangle, 0 the centre of the sphere. Let the tangent at A to the arc AC meet OC produced at E, and let the tangent at A to the arc AB meet OB produced at D; join

; ED. Thus the angle EAD is the angle A of the spherical triangle, and the angle EOD measures the side a. From the triangles ADE and ODE we have

DE?= ADP + AE 2AD, AE cos A,

DE? = OD? + OE? 200.0E cos a; also the angles OAD and OAE are right angles, so that ODP=0A? + ADR and OE = 049 + AE'. Hence by subtraction

we have

0 = 20A? + 2 AD. AE cos A 200.0E cos a ;

ОА ОА AE AD therefore

cos A ; OE OD OE OD

COS a =

+

that is

cos a = cos 6 cos C + sin b sin c cos A.

cos a cos 6 cos c

sin b sin c

Therefore

COS A

We have supposed, in the construction of the preceding Article, that the sides which contain the angle A are less than quadrants, for we have assumed that the tangents at A meet OB and OC respectively produced. We must now shew that the formula obtained is true when these sides are not less than quadrants. This we shall do by special examination of the cases in which one side or each side is greater than a quadrant or equal to a quadrant.

(1) Suppose only one of the sides which contain the angle A to be greater than a quadrant, for example, AB. Produce BA and BC to meet at B'; and put AB' = c', CB' = a'.

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Then we have from the triangle AB'C, by what has been already proved,

cos a' = cos b cos c' + sin b sin c' cos B'AC; but a' =- a, c' = -C, B'AC = 1 A; thus

cos a = cos 6 cos C + sin b sin c cos A.

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(2) Suppose both the sides which contain the angle A to be greater than quadrants. Produce AB and AC to meet at A'; put A'B=(, A'C ='; then from the triangle ABC, as before,

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+

cos a = cos b' cos c' + sin b' sin c' cos A';
Lut W' =-6, c' = 77 - C, A' = 4; thus
l

?A
cos a = cos 6 cos C + sin b sin c cos A.

- T

T. S. T.

C

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