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Subtract the upper from the lower.
Differences 30° 28' 11"
Since we know now all the parts of the triangle except the side p opposite the angle P, that might be found by the proportion, (Art. 64,) 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 one of the second set of Napier's analogies, inverting it, thus ar.comp.log.cos.(s-s)or49°47'30" log.cos.(ss)or80°15'41"
sum rejecting 109.561068=log.tan.p
p=20° and p=40° the distance, between the celestial objects, required.
A method of determining the unknown side at once, if that be the only part required when two sides and the included angle are given, is exhibited at Art. 126.
If two angles and the interjacent side of a spherical triangle were given, the method of proceeding would be to employ the second set of Napier's analogies in the same manner that we have the first in the example above.
The student may take the following example as an exercise: Given the declination of a planet, the angle which its path makes with the meridian through it, and the difference of siderial† time between its passage across a given meridian at the time the declination was observed and some subsequent epoch, to find its declination at the latter epoch and the distance that it has moved.
*This log. need not be found from the tables, but may be obtained by subtracting the ar. comp. of sin (s'+s) above from ar. comp. cos (s'† s) and adding 10; for since tan. 10+ log. sin log. cos. Call
log. tan. =
ing c the ar. comp. log. cos. and s ar. comp. log. sin, this becomes 10+ (10 — s)
− (10 — c) = 10+ c— s. Q. E. d.
+ Siderial time is measured by the daily motion of the fixed stars, and does not vary like solar time with the annual motion of the sun.
Let p in the last diagram be the pole of the equator ; s the first position of the planet, and s' the last ; Ps or s' will be the given codeclination ; P the given difference of time, and s the angle which the planet's path makes with the meridian ; PS or s will be the codeclination, and ss' or p will be the distance moved.
Let P= 2 hours or 30°, s = 70°, and the interjacent side s' = 50° to find s and p.
88. The formulæ 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 co, it will be evident that these formulæ may, when thus applied, be much simplified.
The student can easily make the substitutions necessary to change the foregoing formulæ 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 formulæ which result will be found capable of being expressed in two short rules, or these indeed may be united into a single one.
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
1. The base.
2. The perpendicular. And
3. The complement of the hypothenuse.
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 base and the hypothenuse; and the angle at the vertex is that included between the hypothenuse and perpendicular.
The circular parts are then the parts 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 parts the complements are used.
There being five of these circular parts, it is evident that any three of these which you choose to select will either be contiguous, or else two will be adjoining and one will be separated from them by a part on each side.
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.
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 adjacent 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.
89. Given the sun's right ascension and declination to find his longitude.
Let the parts of the right angled spherical triangle Eas repre
E sent the same circles of the celestial sphere, as at Art. 80, calling the sides opposite the angles by the small letters of the same name with those at the angles, s will be the right ascension, e will be the declination, and q the longitude. sand e are given and q is required.
Of these three parts s, e, and q, s and e are contiguous, and 9 is separated from them by a part on each side; therefore
9 is the middle part and s and e are opposite parts. Applying Napier's rule, remembering that the complement of the hypothenuse q is to be employed, we have
R X sin of comp. q = COS s cos e
Suppose the sun's right ascension, on the 17th May, to be 530 38' and his declination at the same time 19° 15' 57", then
cos 53° 38' x cos 19° 15' 57" 9
1010 ar. comp. log. 1010 = 0.000000
log. cos. 53° 38' 9.773018 log. cos. 19° 15' 57" 9.974971
sum rejecting 10 9.747989 = log. cos. 550 57' 43' Hence q or the longitude required is 55° 57' 43".
90. The same being given as in the last example, required the obliquity of the ecliptic. The required part is the angle e in the figure. Of E, s and
E e, since the three are contiguous leaving out the right angle, s is the middle part, hence applying the rule of Napier,
R sin s = tan e çot 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
R sin s cot E=
tan e Applying logarithms to the second member, we have ar. comp. log. tan e 19° 15' 57" = 0.456521 log. sin s 53° 38'
9.905925 log. R
sum rejecting 10= 10.362446 = CO E
E= 23° 27' 50%* 91. Had the angle s been required, it might be found by observing that of the three parts, s, e and s,the latter two of which
• The obliquity of the ecliptic is continually, though very slowly, changing. It is always given with the minutest attainable accuracy in the Nautical Almanaa