are given, e is the middle part, and s and s adjacent. Hence, using the complement of the oblique angle si i sin e= tan s cot s whence i sin e cot s tan s Applying logarithms as before, the value of s will be found. We leave this as an exercise for the learner. In the solution of the above triangle, it will be observed that we have found each of the unknown parts in terms of the two given, and have not employed one of those first calculated to obtain another. This is agreeable to the principle laid down at Art. 41, of plane trigonometry, and the reason is the same. Such a method of proceeding is always practicable in the solution of right angled spherical triangles. In the examples which we have taken above, we have supposed the base and perpendicular of a right angled spherical triangle given. Any other two parts being given, each of the unknown parts may be calculated by the aid of Napier's rules, in a manner entirely similar to what has been just exhibited. 18 EXAMPLE III. 92. Given the sun's declination to find the time of his rising and setting at any place whose latitude is known. Let nEs a represent the meridian of the place, z being the zenith, and нo the horizon', and let s's" be the apparent path of the sun on the proposed day, cutting the horizon in se Then the arc EZ will be their ho latitude of the place, and consequently EH, or its equal qo, H E S Z R will be the colatitude, and this measures the angle oAQ; also RS will be the sun's declination, and AR, expressed in time, will express the time of sunrise from 6 o'clock, for nas is the six o'clock hour circle. Hence, in the right-angled triangle SAR, we have given RS and the opposite angle a to find AR, the time from 6 o'clock. EXAMPLE I. Required the time of sunrise at latitude 52° 13′ N., when the sun's declination is 23° 28'. By Napier's rule, Rad. sin AR cot A. tan RStan. lat. tan. dec. 3 43 46 35 time of rising. * Degrees are converted into hours by multiplying by 4 and dividing by 60, which is equivalent to multiplying by 15. SCHOLIUM. It should be here remarked that the time thuis determined is apparent time, which is that which would be shown by a clock so adjusted as to pass over 24 hours during one apparent revolution of the sun, or from its leaving the meridian to its return to it again, the index pointing to 12, when the sun is on the meridian. But it is impossible that any clock can be so adjusted, because the interval between the sụccessive return of the sun to the meridian is continually varying, on account of the unequal motion of the sun in its orbit, and of the obliquity of the ecliptic; each of these varying intervals is called a true solar day, and it is the mean of these during the year which is measured by the 24 hours of a well regulated clock, this period of time being a mean solar day; hence, at certain periods of the year, the sun will arrive at the meridian before the clock points to 12, and at other periods the clock will precede the sun; the small interval between the arrival of the index of the clock to 12 and of the sun to the meridian is called the equation of time, and it is given in page ii. of the Nautical Almanac for every day in the year; this correction, therefore, must always be applied to the apparent time determined by trigonometrical calculation to obtain the true time, or that shown by a well regulated clock or chronometer. Another circumstance too must be taken into account, in order to determine the apparent time with rigorous accuracy, viz. the change in the declination of the sun from sunrise to noon. In the Nautical Almanac, the declination of the sun is given for every day at noon, and if this be used in the computation we shall assume that the declination has not varied from sunrise to noon, which is not the case; hence it will be necessary to compute the declination for the time of sunrise, as determined above, and then to resolve the problem with this corrected declination. The correction is obtained by taking from the Nautical Almanac the variation of declination in 24 hours, and then finding by proportion the variation for the time required 2. Required the time of sunrise at latitude 57° 2' 54", when the sun's declination is 23° 28'? 3' 11' 49" PROBLEM III, 93. Given the latitude of the place, and the declination of a heavenly body, to determine its altitude and azimuth when on the six o'clock hour circle. the right-angled triangle SBA, the given quantities are as, the declination, and the arc op, or angle SAB, the latitude of the place, to find the altitude вs, and the azimuth Bо from the north point of the horizon; or to find the complement AB of this azimuth, that is, the sun's bearing from the east. EXAMPLES. 1. What was the altitude and azimuth of Arcturus, when upon the six o'clock hour circle of Greenwich, lat. 51° 28' 40" N., on the 1st of April, 1822; its declination on that day being 20o 6' 50" N.? By Napier's rule we have Rad. sin Bssin a sin As Rad. cos A Rad. cos Atan AB cot As...* cot Bo cot As * This sign.. signifies therefore. For the altitude. For the azimuth. sin A 510 28' 40" 9.893410 COS A 9.794361 sin as 20 6 50 9.536416 cot AS 10.436255 sin BS 15 36 27 9.429826 cot BO 770 9'4" 9.358106 Hence the altitude is 150 36' 27", and the azimuth 770 9'4"N. .94. There remains one case in the solution of oblique angled spherical triangles, which we have deferred to this place, because we wished to employ in it the rules for the solution of right angled triangles. This is where two sides and the angle opposite* to one of them, or two angles and the side opposite to one of them are given ; or as it is sometimes expressed, where two of the given parts are a side and its opposite angle. In such a case we may proceed as follows, By means of the proportion, the sines of the angles are as the sines of the opposite sides (Art. 81), the unknown part opposite one of the given parts may be found. Four parts of the triangle will then be known, and two will remain unknown; these two will be a side and its opposite angle, to find which, from the vertex of the unknown angle let fall an arc perpendicular upon the unknown side opposite, and the given triangle will be divided into two partial triangles, which will be right angled and in each of which two parts will be known. Applying Napier's rules to the solution of these, the partial angles which compose the unknown angle may be found, and their sum will be the value of the unknown angle; then the unknown side opposite may be found by the proportion, the sines of the angles are as the opposite sides; or this last side may be found by calculating the two parts of which it is composed from the right angled triangles, and adding them together, which is the better method since it avoids ambiguity. If the perpendicular arc, drawn from the vertex of the unknown angle to the unknown side falls with * The term opposite is used in a niore exact sense here than in Napier's rules, of which we have just been speaking. |