METHOD II. From the sum of the three sides take the side om to the angle sought; and add the arithmetical complements of the se of the two containing sides, and the sines of the sum and remane and the sum of these four is the cosine of the angle sought.

METHOD III. Take the sum and difference of the base, and t difference of the sides, and then add the sines of this sum and differen and the arithmetical complements of the sines of the containing side and the sum of these four is the sine of the angle sought.

Norr. Instead of taking the sum and difference of the base, the difference of the sides, the two containing sides may be subtrada from the sum of the three sides.

METHOD IV. Add the arithmetical complements of the sines of th half sum and of its excess above the base, and the sines of its excesses above the other two sides; and the sum of these four is the tange of the angle sought.

NOTE. In using the common tables of logarithms, the third meth is more accurate than the second when the required angle is small, the second is more accurate when it is large. The fourth method may be used in all cases, except when the angle sought is very nearly equi to two right angles.

94 30 43 2 Angle ACB.

2. Given the sides AC 50° 54' 32′′, CB 37° 47′ 18′′, AB 74° 51′ 50′′; to find the angles. Ans. The angles at 44° 10′ 40′′, at A 33° 22′ 45′′, and at C 119° 55′ 6′′.

and

3. Given AB 58° 0′ 5′′, AC 88° 12′ 28′′-8, and BC 94° 52 40"-8; to find the angles. Ans. The angles at C 57° 40 21′′-6, at B 84° 49′ 2′′, and at A 96° 33′ 28′′.

CASE VI. WHEN THE THREE ANGLES ARE GIVEN.

1. In the oblique-angled spherical triangle ABC are given the angles A 78° 25', B 110° 30′, and C 64° 48′; to find the side

B

nstruction. Draw the radius e primitive, and make OF the -tangent of 69° 30′-180°— '30', and make OG its tangent, with its secant describe from e circle BCE. Lay the semient of 134° 18′ 69° 30'+64° rom O to H, and the semi-tanof 4° 42′ 69° 30′-64° 48' 10 the same way to K, and a the diameter HK describe circle HPK, and with the se

angent of 78° 25' from O cut that circle in P. Join OP, and on y OM the tangent of 78° 25', and with the secant of 78° 25' from lescribe the circle ACD; then ABC is the triangle.

Describe a great circle through the points F and P. The triangle P is semi-supplemental to ABC. For OP is the measure of BAC, use O and P are the poles of AB and AC; FP is the measure ACB, because F and P are the poles of BC and CA, and OF is the plement of ABC. Also, AB is the measure of the angle POF, bese A and B are the poles of PO and OF; and BC is the measure of 'P, because B and C are the poles of OF and FP, and AC is the plement of OPF.

Calculation. To find BC or the angle OFP. Take OF the supplent of ABC 69° 30', and the difference between it and C or PF is 4° and the half of it taken from BAC or OP, and added to it, are 51 and 41° 33'. Add the arithmetical complements of the sines CBD and of C, and the sines of 41° 33' and of 36° 51'; and the n of these four is the sine of OFP or of BC. In the same man, we find AB and AC.

2. Given the angles A 44° 10′ 40′′, B 33° 22′ 45′′, and C 19° 55' 6"; to find the sides. Ans. The sides BC 50° 54' 08, AB 74° 51′ 46′′-3, and AC 37° 47′ 17′′-5.

3. Given the three angles A 87° 46′ 13′′, B 46° 34′ 5′′ nd C 53° 39′ 20′′; to find the sides. Ans. The sides AB 31° 4', BC 40° 16′, and AC 28° 1′.

APPLICATION OF SPHERICAL TRIGONOMETRI TO THE SOLUTION OF ASTRONOMICAL PROBLEMS

SPHERICAL TRIGONOMETRY is of great use in Astronomy, Geography, and Navigation; and therefore a few example of its application to these sciences are given here, after explaising the circles of the sphere.

To lay down the circles of the sphere on the plane of the meridian of Edinburgh, in Lat. 55° 57′ 20′′ N.

Let the primitive be the meridian. Draw the diameter HR for the horizon, and the perpendicular diameter ZN; then Z is the zenith, and N the nadir. Make RP, ZE, each 55° 57′ 20′′, and draw the diameters Pp and EQ; then P and p are the poles, and EQ the equator, and Pp the hour-circle of six. About the points P and p as poles describe the circles def and g h k for the polar circles at the distance of 23° 28′; and in the same manner describe the tropics about the poles P and p at the distance of 66° 32'.

Suppose the time for which the circles are drawn to be the 3d A gust, 1831, at 9h 36m in the morning. The declination for that time is 17° 40′ N, About the pole P, at the distance of 72° 20′, describe the circle a Cb, which is the parallel of the sun's declination for that day. Let it meet HR in A, Pp in C, and ZN in F, and describe the great circles PAP, PFp, meeting EQ in B, G, and ZCN meeting H in D. Describe the great circle PSP, making the angle ZPS 36° -2 24m the time from noon, and describe the circle ZSN meeting HR in T, and let PSP meet EQ in M.

The point b is the sun's place at midnight, and a his place at noon; A the point where he rises, C is his place at six, F his place when due east, and S his place at the given time. The circle ZON is the prime, or east and west vertical circle; O the east or west point of the horizon, Rits north, and H its south points; Rh is the sun's depression & midnight, aH is his meridian altitude, ST his altitude at the given time, OF his altitude when east, and CD his altitude at six. The arch QB, or the angle QPB, is the time of the sun's rising from midnight, and BO or BPO the time from six, which is called the sun's ascensional difference; BE, or BPE, the time of his rising from noon; OG, or OPG, the time from six, when he is due east; and GE, or

'E, the time from noon.

Also OM, or OPM, is the given time n six, and EM, or EPM, the given time from noon. AR, or R, is the sun's amplitude from the north; OA, or OZA, his amude from the east; and AH, or AZH, from the south. RD, or D, is his azimuth from the north at six; and DH, or DZH, from south. And HT, or HZT, is his azimuth from the south at the en time; and TR, or TZR, that from the north.

PROBLEM I. Given the obliquity of the ecliptic, and the n's longitude, to find his declination and right ascension. In the spherical triangle SMO, right-angled at M, are given the anSOM, the obliquity of the ecliptic, and the side SO the sun's lonude; to find SM the declination, and OM the right ascension. These › found by Case 1. Right-angled Triangles.

PROBLEM II. Given the latitude of the place, and the n's declination, to find his amplitude, and the time of his sing.

In the spherical triangle APR, right-angled at R, are given PR the titude, and the hypotenuse PA, the polar distance = 90° the n's declination; to find AR the amplitude from the north, and e angle APR, which, converted into time at the rate of 15° to an ›ur, gives the time from midnight when the sun rises. Wherefore, by ase 4. of right angled spherical triangles, cos. Latitude: R:: cos. ›lar dist. or sin. decl. cos. RA, or sin. OA, the amplitude.

And tan. polar dist. : tan. Lat. :: R: cos. P, the time of rising. The same things may be found in the triangle OAB right-angled at where AOB or RQ is the co latitude, and AB the declination; to nd AO the amplitude, and OB the ascensional difference, which, subacted from six hours, gives the time of sun-rising. This is wrought y Case 3.

PROBLEM III. The same things being given, to find the un's azimuth and altitude at six o'clock.

In the triangle PCZ, right-angled at P, are given PZ the co-latitude, nd PC the polar distance; to find ZC the zenith distance, or complenent of CD the altitude, and the angle CZP the azimuth from the orth. Wherefore, by Case 5. R: cos. ZP = sin. Lat. cos. CP = sin. declination: cos. CZ, or sin. CD the altitude.

And sin. ZP = =cos. Lat. : R:: tan. PC, or cot. decl.: tan. Z, the

zimuth.

The same things may be found in the triangle OCD, right angled at D, where COD or PR is the latitude, and OC the declination. s wrought by Case 1.

This

PROBLEM IV. The latitude and declination being still given, to find the sun's altitude, and the time when he is east.

In the triangle ZPF, right-angled at Z, are given ZP the co-latitude, and PF the polar distance; to find ZF the zenith distance, and the an gle ZPF the time from noon. Wherefore, by Case 4. cos. ZP, or sin. Lat. R cos. PF, or sin. decl. : cos FZ, or sin. FO, the altitude. And tan. FP, or cot. decl. : tan. ZP, or cot. Lat. :: R: cos. P.

The same things may be found in the triangle FOG, right-ang G, in which are given FOG, or ZE, the latitude, and FG the deci tion; to find FŐ the altitude, and OG the complement of GE, time from noon. This is wrought by Case 3.

PROBLEM V. Given the latitude, declination, and hour; w find the sun's altitude and azimuth at that time.

In the triangle OSM, right-angled at M, are given MS the decline tion, and MO the time from six, to find the angle MOS (by Case 5 sin. OM: R:: tan. MS: tan. O, and SOM + colat. EOH = 801. Also R: cos. MO:: cos. MS: cos. SO. Then in the triangle 051. right-angled at T, are given SO, and the angle SOT; to find 01 the complement of TH, the azimuth, and TS the altitude, by Case 1.

The same things may be found by resolving the oblique-angled tangle PZS, in which are given PZ the co-latitude, PS the polar dis tance, and the angle ZPS the hour from noon; to find ZS the rent distance, and the angle at Z the azimuth, by Case 3. of oblique-angle spherical triangles.

PROBLEM VI. Given the latitude and longitude of the moon, or of a star, and the obliquity of the ecliptic; to find the right ascension and declination.

Suppose HR the equator, and EQ the ecliptic, then the latitude of the moon or any star S is MS, the longitude OM, the right ascension Of the declination TS, and the obliquity of the ecliptic TOM. Therefore in the triangle OMS, right-angled at M, are given the two sides O and MS about the right angle, to find the side OS and the angle MOS which are found by Case 5. Now it is evident that when the moon or star S is without the ecliptic, the angle MOS added to the obliquity f the eliptic will give the angle TOS, or when S is within the ecliptic the difference of these angles will be TOS. Hence in the triang OTS, right angled at T, are given the angle TOS, and the hypotenust OS, to find the sides OT and TS, which are done by Case 1.

PROBLEM VII. Given the latitude of the place, and the sun's declination; to find the time when twilight begins and ends This problem is solved by Case 5. Oblique-angled Triangles, since there are given the polar distance, the co-latitude, and the zenith dis tance === 90° 18°, which form the three sides of an oblique-angled spherical triangle, from whence to find the angle at the pole opposite the zenith distance, which is the time from noon that twilight begins and ends.

PROBLEM VIII. Given the right ascensions and declina tions, or the longitudes and latitudes of two celestial objects ; to find their distance.

This problem is solved by Case 3. Oblique-angled Triangles, since there are given two sides and the contained angle to find the opposite side. The sides are the complements of the declinations, or latitudes and the contained angle the difference between the right ascensions, or longitudes. By this problem the distances of two places on the globe may be found, of which the latitudes and longitudes are given; for the