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62. The formulæ requisite for the solution of right-angled triangles may be obtained from the preceding Chapter by supposing one of the angles a right angle, as C for example. They may also be obtained very easily in an independent manner, as we will now shew,

С

B

Let ABC be a spherical triangle having a right angle at C; let O be the centre of the sphere. From any point P in OA draw PM perpendicular to OC, and from M draw MN perpendicular to OB, and join PN. Then PM is perpendicular to MN, because the plane AOC is perpendicular to the plane BOC; hence

PN = PM? + MN= 0PS OMR + OM? ON? = 0P- ON; therefore PNO is a right angle. And

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ON ON OM
OP OM OP

that is, cos c= cos a cos b

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PM PM PN

that is, sin b = sin B sin cl
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Similarly sin a = sin A sin c
MN MN PN

that is, tana=cos B tanc cl
ON PN'

Similarly tan =cos A tan c
PM PM MN
OM MN' OM' that is, tan b=tan Bsina)

Similarly tan a=tan A sin b

(3),

.(4).

T. S. T.

D

Multiply together the two formulæ (4); thus,

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Multiply crosswise the second formula in (2) and the first in (3); thus sin a cos B tan c = tan a sin A sin c;

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These six formulæ comprise ten equations; and thus we can solve every case of right-angled triangles. For every one of these ten equations is a distinct combination involving three out of the five quantities a, b, c, A, B; and out of five quantities only ten combinations of three can be formed. Thus any two of the five quantities being given and a third required, some one of the preceding ten equations will serve to determine that third quantity.

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63. As we have stated, the above six formulæ may be ohtained from those given in the preceding Chapter by supposing C a right angle. Thus (1) follows from Art. 39, (2) from Art. 41, (3) from the fourth and fifth equations of Art. 44, (4) from the first and second equations of Art. (44), (5) from the third equation of Art. 47, (6) from the first and second equations of Art. 47.

Since the six formulæ may be obtained from those given in the preceding Chapter which have been proved to be universally true, we do not stop to shew that the demonstration of Art. 62 may be applied to every case which can occur; the student may for exercise investigate the modifications which will be necessary when we suppose one or more of the quantities a, b, c, A, B equal to a right angle or greater than a right angle.

64. Certain properties of right-angled triangles are deducible from the formula of Art. 62.

From (1) it follows that cos c has the same sign as the product cos a cos b; hence either all the cosines are positive, or else only one is positive. Therefore in a right-angled triangle either all the three sides are less than quadrants, or else one side is less than a quadrant and the other two sides are greater than quadrants.

From (4) it follows that tan a has the same sign as tan A. Therefore A and a are either both greater than or both less

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than ä; this is expressed by saying that A and a are of the same

2 affection. Similarly B and b are of the same affection.

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65. The formulæ of Art. 62 are comprised in the following enunciations, which the student will find it useful to remember ; the results are distinguished by the same numbers as have been already applied to them in Art. 62; the side opposite the right angle is called the hypotenuse : Cos hyp = product of cosines of sides

.(1),
Cos hyp = product of cotangents of angles ....... ...(5),
Sine sider sine of opposite angle x sine hyp .... .(2),
Tan side = tan hyp x cos included angle

(3),
Tan side = tan opposite angle x sine of other side...... (4),
Cos angle= cos opposite side x sine of other angle...... (6). .

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66. Napier's Rules. The formulæ of Art. 62 are comprised in two rules, which are called, from their inventor, Napier's Rules of Circular Parts. Napier was also the inventor of Logarithms, and the Rules of Circular Parts were first published by him in a work entitled Mirifici Logarithmorum Canonis Descriptio...... Edinburgh, 1614. These rules we will now explain.

The right angle is left out of consideration; the two sides which include the right angle, the complement of the hypotenuse, and the complements of the other angles are called the circular

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parts of the triangle. Thus there are five circular parts, namely, a, b, 3-4, 5-c, -B; and these are supposed to be ranged ,

, 2

2 round a circle in the order in which they naturally occur with respect to the triangle.

Any one of the five parts may be selected and called the middle part, then the two parts next to it are called adjacent parts, and the remaining two parts are called opposite parts. For example, if-B is selected as the middle part, then the adjacent

2 parts are a and - c, and the opposite parts are b and – .

A. 2

2 Then Napier's Rules are the following: sine of the middle part = product of tangents of adjacent parts, sine of the middle part = product of cosines of opposite parts.

67. Napier's Rules may be demonstrated by shewing that they agree with the results already established. The following table shews the required agreement : in the first column are given the middle parts, in the second column the results of Napier's Rules, and in the third column the same results expressed as in Art. 62, with the number for reference used in that Article.

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C

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sin

= cos a cos 6

cos c = cos a cos b..(1),

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2

B

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- C

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-o sin (5 –c) = tan (5 – 4 ) tan (5 – B)

-(1-0)-
1-B sin (-) = tan a tan (-)

B
sin (-B) = cos bcos (-4)
6
sin a = tan b tan (-)

-B
(-4) cos (1–0)
(5-1)

(6-B) cos (7 –) 3-4 sin (-4) = tan otan (6-6)

c ain (-1) = cos a cos (-)

-B

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The last four cases need not have been given, since it is obvious that they are only repetitions of what had previously been given; the seventh and eighth are repetitions of the fifth and sixth, and the ninth and tenth are repetitions of the third and fourth.

68. It has been sometimes stated that the method of the preceding Article is the only one by which Napier's Rules can be demonstrated; this statement, however, is inaccurate, since besides this method Napier himself indicated another method of proof in his Mirifici Logarithmorum Canonis Descriptio, pp. 32, 35. This we will now briefly explain.

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