site the angle P, that might be found by the proportion (Art. 81), the sines of the angles are as the sines of the opposite sides. But to avoid an ambiguity in the result similar to that of Art. 81, and the trouble of determining which of the two results corresponds to the other parts of the triangle now fixed, it is better to employ XI. of Napier's analogies, inverting it as seen in the last column above. This gives p= about 20° and p = about 40° = 240 geographical miles, the distance required.

=

EXAMPLE II.

Given the moon's R.A. 11" 38" 27*•15.

Dec. 4° 5' 40" 4 N.

To find her latitude and longitude, the obliquity of the ecliptic being

87. When, in the case considered in Art. 86, the only part required happens to be the side opposite the given angle, the finding of the other two angles then becomes merely a subsidiary operation, and the determination of the required side, by Napier's

* The summer solstice is 90° from the vernal equinox, and is N of the equator. The pole of the ecliptic p' therefore will be south of P and between it and the 2700 point. The right ascension of the moon being nearly 12 hours is nearly 1800, which fixes its place in the diagram.

The determination of the latitude and longitude of a heavenly body from its right ascension and declination as above is one of the most useful problems in Astronomy, the right ascension and declination being observed directly with the astronomical instruments as will be explained in a subsequent part of the work, and the latitude and longitude being required for computing the elements of the orbits of the heavenly bodies.

analogies, seems unnecessarily long. A shorter method of solution is deducible from the fundamental formula, obtained at Art. 82, or

cos ccos a cos b+ sin a sin b cos c

For substituting cos a tan a for its equal sin a it becomes

cos ccos a (cos b+tan a sin b cos c)

(1)

Hence, to find the side c, we must determine a subsidiary angle w from the

=

1. In a spherical triangle are given a 38° 30′ b⇒ 70°, and c = 31° 34′ 28′′,

88. If when two angles and the included side are given, the angle opposite to the given side be the only part required, a similar formula should be employed, deduced as follows. From the fundamental formula (1) above, may be obtained by aid of the polar triangles, the formula

cos CCOS A COS B+ sin a sin B cos C.

which becomes when cos A tan A is substituted for sin a,

COS A sin

sin w

Hence, having found a subsidiary angle w, by the equation

89. The formulas for the solution of spherical triangles in general, which have now been demonstrated, apply of course to right angled triangles; but if it be recollected that the trigonometrical lines of the right angle or 90° are either R, 0, or ∞, it will be evident that these formulas may, when thus applied, be much simplified.

The student can easily make the substitutions necessary to change the foregoing formulas into such as apply exclusively to right angled triangles, for himself. We shall not occupy space with them here, but be content with observing that after they have been made, all the formulas which result will be found capable of being expressed in two short rules, or these indeed be united into a single one. may

*

Amongst all the convenient and useful inventions of mathematicians, none is more ingenious and beautiful than this, the author of which is the celebrated Lord Napier, whose name we already have had occasion repeatedly to mention in connection with the most happy discoveries for facilitating mathematical operations. The rules are known as

NAPIER'S RULES FOR THE CIRCULAR PARTS.

The circular parts of a right angled spherical triangle are

The two sides including the right angle, called

And

1. The base.

2. The perpendicular.

3. The complement of the hypothenuse

4. The complement of the angle at the base.

5. The complement of the angle at the vertex.

The right angle being entirely left out of consideration in the solution of triangles of this kind, the angle at the base is that included between

The mode of deducing them is given in App. II. p. 194.

the base and the hypothenuse; and the angle at the vertex is that included between the hypothenuse and perpendicular.

The circular parts are then the elements of the triangle itself except the right angle, only that the complements of the hypothenuse and oblique angles are the circular parts, instead of these themselves.

The annexed diagram shows of which elements of the triangle the complements are used.

There being five of these circular parts, it is evident that any three of them which you choose to select will either be contiguous or else two will be contiguous, and one will be separated from them by a part on each side.

Comp

Comp

Comp

In the first case, the part intermediate between the other two is called the middle part, and they are called its adjacent parts. In the second case, the part which is separated from the other two is called the middle part, and they its opposite parts. By means of this arrangement, all the relations of a right angled triangle may be expressed in the two following rules of Napier :

1. Radius multiplied by the sine of the middle part is equal to the rectangle of the tangents of the adjacent parts.

2. Radius multiplied by the sine of the middle part is equal to the rectangle of the cosines of the opposite parts.

=

Or both rules may be given thus: radius into the sine of the middle part the rectangle of the tangents of the adjacent parts the rectangle of the cosines of the opposite parts.*

The memory will be aided by observing that the words tangents and adjacent in the second clause of the above rule both contain the letter a; and that the words cosines and opposite in the last clause both contain the letter o.

As the right angle of a right angled spherical triangle is always known, any other two parts being given, the rest may be found by the above rules.

The method of proceeding is as follows: Take the two given parts and one of the required parts, or if but one of the unknown parts be required, take that, you will thus have under consideration three parts of the triangle. One of these three will be middle, and the other two either adja

* The rule may be read without radius which must be understood as entering the resulting formulas, in accordance with the principles of homogeneity.

cent or opposite; apply the rule of Napier, and you will have an equation resulting which will contain the two given parts and the required part; make the required part the unknown quantity in the equation, and resolve it, you will thus obtain the value of the required part in terms of the two which were given. By applying logarithms to this value, you will have it in degrees, minutes, and seconds.

EXAMPLE

90. Given the sun's right ascension and declination to find his longitude.

Let the parts of the right angled spherical triangle EQS represent the same circles of the celestial sphere, as at Art. 80; a and & are given, and 7 is required.

Of these three parts a, & and 1, a and

E

S

are contiguous, and 7 is separated from them by a part on each side; therefore is the middle part and a and are opposite parts. Applying Napier's rule, remembering that the complement of the hypothenuse is to be employed, we have

91. The same being given, required the obliquity of the ecliptic. The required part is the angle E in the figure. Of E, a and 8, since the three are contiguous leaving out the right angle, a is the middle part, hence applying the rule of Napier,

sin a tan & cot E

We put cot E instead of tan E, because, according to the directions before given, the complements of the oblique angles are to be employed. Taking the value of cot E from the above equation, we have

The sun's R.A on 1st of May 1850 is 2′ 33′′ 10*42, and his declination at the same time 15° 3' 2"-4 required his longitude and the obliquity of the ecliptic.