: sin. Asin. B:: 2 sin. by geometry A D + DC·AD—DC=AB+ Forc:a-b:: sin. C: sin.A- sin. B:: sin. A+ B BC AB-BC; and, as AD+DC=AC, we have A CAB+ BC::AB-BC: AD-DC. Formula 5.—In any triangle, as ABC (fig. 2 cos. C A-B A : COS. sin. 2 2 But A+ B - 2 A+ B A+ B : 2 2 For ca+b:: sin. C: sin. A + sin. B. COS. 2 (formula 4, prop. III.), and sin. A + sin. B = 2 sin. A+ B 2 A 2 B (formula 3, prop. III.); therefore c: a + b :: 2 sin. 2 : COS. 2 Formula 7.. In any plane triangle, as ABC (figs. 8 and 9), cos. CAB= - C B2 2 СА . А В' radius being unity. For in fig. 8, C B2= C A2 + A B2 + 2 CA. AD; A+ B COS. : 2 A+ B COS. : : 2 A+ B C proportion becomes c : a+b:: sin. A-B 2 2 CA.AB' C B2 + 2 CA. AD; and AD=AB. the cos. A; therefore C A2 + A B C B2 + 2 CA. Formula 6.-In any triangle, as ABC C A+B (fig. 10), ca — b: cos. : sin. 2 2 AB cos. A; therefore cos. A= CA+A B2C B 2 CA.AB' or, calling the sides oppo site the angles A, B, and C, respectively a, b, and c, Formula 8.-In any triangle, as ABC (fig. 10), if Examples of the application of the formula for the solution of right angled plane triangles. 1. In a plane right angled triangle, as ABC', fig. 11, given AC 480, and ▲ A 53′ 8', required AB and BC. By Prop. VII. AB= AC sin. A ; these equations solved logarith Ꭱ AB sin. B sin. C Examples of the application of the formula for the solution of oblique angled plane triangles. 1. In a triangle, as ABC, fig. 10, given A B, 376, ▲ A 48° 3′, and B 40° 14′, to find the other points. The angle C, being the supplement of the sum of A and B, is 91° 43′. whence A C = R2 sin. A BC sin. C, AB sin. A ; and B C = sin. C 2. In a triangle ABC, fig. 12, are given BC 2943, A C 2314, and given sides, 139024, to find the other parts. Bopposite the less of the This example belongs to what has been denominated the ambiguous case in oblique angled triangles. For a circle described from C, with the radius CA, will cut the base again in A', and both the triangles BCA and BC A' answer to the given data; and this conclusion agrees also with Log. BC, 2943 sin. B 39° 24' 3.468790 Log. AC, 2314 13.271379 that obtained from the formula for finding A, which is sin. A = BC sin. B ; whence the result may be the angle A or its supplement; or, in the figure, the result may be the angle A, or its supplement BA'Č. The supplement of the sum of B and A is BCA = 86° 46′ 14′′; and the supplement of the sum of B and BA'C is BC sin. BCA · cosec. A BCA 14° 25′ 46′′ and BA= R2 BC sin. BCA' cosec. BA'C and B A' = R2 BA 3639-8 log. 20 R2 20 3.561084 3. In a triangle, as ABC, fig. 9, given AC 2847, BC 979-3, and the angle C included by those sides 57° 21' 10'', to find the other parts. = From B demit BD a perp. on AC; then CD = CA-CD. CB. cos. C prop. VII; and AD = Then, Formula 2, proposition VIII., CD: DA:: cot. C: cot. A, or cot. A = the angle A B C is the supplement of the sum of A and C; and A B = BC sin. C cosec. A is given, and by the formula referred to AC +CB: AC-CB: tan. The angles may be found more elegantly and conveniently by means of formula 3, Prop. VIII. For as the angle C is given, its supplement or the sum of the angles A and B is given; consequently A+ B 4. Given A C (fig. 9), 4287, A B, 3758, and BC 2819, to find the angles. We shall, for the sake of exemplifying the use of different formulæ, find each angle by a different process. S By Formula 4, Prop. VIII, we have AC: AB+BC: AB-BC: AD DC', whence AD-DC AB+ BC AB · ВС and cos. C = 2 CD rad. CB . 2AC and C D = AC 2 AD-DC 2 Now A B+ BC 6577, A B-BC 939, and 2 AC 8574, whence whence CD= 1423.24. Putting S for half the sum of the three sides, and denoting the sides opposite the angles A, B, and C, by a, b, and c respectively, we have by Form. 8, Prop. VIII., cos. rad2 S b.c S 3. At D (fig. 15), I took BDC the angle of elevation of a light-house on the top of a hill 43° 24', and measuring directly towards it 180 yards, to C, I took again the angle of elevation B C E of its top, 63° 12′, and A C E the elevation of its bottom 49° 14', required the height of the tower, and also the height of the hill on which it stood? Here, as in the last example, we have DB C= BCE-BDC=19° 48', and BE DC sin. D cos. DBC sin. DCE rad.3 326-03, and by Form. 2, Prop. VIII., EB: EA :: tan. BCE tan. A C E, whence E A EB tan. ACE 7. In a tower besieged there are three steeples, A, B, and C (fig. 19), with the distances of which from each other, the besiegers are acquainted from a map of the place; A B is 700, BC 560, and AC 1000 yards. Now the be191-41, siegers wishing to plant a battery at D are desirous of knowing whether from thence the town can be attacked with effect. The angle CD B was found to be 12° 15', and the angle B DA 15° 35'; required the distance of D from each of the objects A, B, and C. 4. To determine the distance of a tower on the opposite side of a river, (see fig. 16) an officer measured a base A B of 346 yards, and at A found the angle CAB 67° 14', and at B the angle C BA 48° 28', what are the distances of C, the tower, from A and B, the extremities of the measured base? The angle A C B being the supplement of the sum of A and B is 64° 18'. Hence AC AB sin. B⚫ cosect C =287-4, and BC= rad.2 5. To find the distance between a house and a mill, neither of which I could get near, I measured a base A B (fig. 17), = 820 yards; at AI observed the angle DAB = 80° 28′, and C A B = 40° 43′; and at B I observed the angle C BA= 100° 12′, and D BA = 52° 46′, required the distance CD? The angle ADB being the supplement of the sum of DAB and A B D is 46° 46'; and the angle A C B being the supplement of the sum of CAB and C BA is 39o 5′; also the angle DAC, which is the difference of D A B and CA B, or 39° 5'. AB sin. ABD cosect ADB Now AD= rad. 2 ADC+ACD 2 =848-444; and 27′ 30′′; and by Form. 3, Prop. VIII., tan. ADC ACD On AC let a segment of a circle be described to contain the sum of the angles CDB and ADB; and on BA imagine the segment of a circle to be described, containing the angle AD B, the intersection of these circles D, as is obvious from the principles of geometry, will be the position of the battery. Join BD and let it meet the circumference in E, then the angle C EA, which is the supplement of CDA, is 152° 10′, and the angle CAE = the angle CDE 12° 15', and ECA= EDA 15° 35'. Hence C E = CA sin. CAE cosect. CEA . rad.2 454·438; and calling BC, a, AC, b, and A B, c; and putting a+b+c S for we have, formula 8, proposition 2 2 C or 2 Crad. SS VIII., cos. a b = 42° 39′ 46′′, and 21° 19′ 53′′; whence A C B consequently BCE=ACB-ECA = 27° 4' 46". Then in the triangle BCE we have BC = 560, C E 454-438, and the angle BCE= 27° 4′ 46′′, to find the angle CBE, which, by operations already sufficiently exemplified, is found= 50° 50' 54". Now in the triangle C B D, wehave given the angle C D B = 12° 15′ and the angle CBE 50° 50′ 54′′, and consequently the angle BCD=116° 54' 6"; and the side BC being given, we have BD: = BC sin. BCD cosect. CDB CD = =4° 24′ 1′′; 2046-7; and hence DA is readily found whence ADC66° 3′ 29′′, and ACD 74° 51' 31"; and CD= AD sin. DAC cosect A CD . rad.2 585.26. 2061.6. = ON SPHERICAL TRIGONOMETRY. PROP. I.-In any right angled spherical triangle, the rectangle of radius, and the sine of either of the sides containing the right angle, is equal to the rectangle of the tangent of the other side, and the cotangent of the angle opposite to that side. Let ABC (fig. 20), be a spherical triangle, right angled at B; and let D be the centre of the sphere. From B, in the plane BAD, draw BE perpendicular to DA, and from E in the plane A CD draw E F also perpendicular to D A, meeting DC produced in F, EF and EB at their point of meeting, it is perand join FB. Then as AD is perpendicular to pendicular to the plane E B F, and therefore the plane BDA, which passes through AD, is also perpendicular to the plane B E F. Now, as the angle ABC is a right angle, the plane BDC is |