: 2 cos. : 2 2 cos. 을 : sin. 2 2 2 2 2 2 2 2 COS. by geometry A D+DC AD-DC=AB + Forc:a–6::sin.C:sin. A --sin. B::sin. A + B BC: AB-BC; and, as AD +DC=AC, : sin. A A + B sin. B:: 2 sin. A +B we have AC: A B + BC :: AB-BC: 2 AD- DC. A + B A-B A + B sin. Formula 5.--In any triangle, as A B C (fig. : ; sin. 2 C A - B 10), c:a + b :: sin. sin. :: cos. 2 For c:Q+6:: sin. C :sin. A + sin. B. But Formula 7.- In any plane triangle, as A+B A + B A B C (figs. 8 and 9), cos. C AB = sin.C=sin. A + B = 2 sin. A+ A B2 – CB? radius being unity. For (formula 4, prop. III.), and sin. A + sin. B = 2 2 CA. A B' A +B A-B fig. 8, C B’ = C A' + A B? + 2C A.AD; sin. (formula 3, prop. III.); and AD = AB . cos. BAD = - AB.cos. 2 A + B A +B CAB, whence C B = CAP + D BP – 2 CA therefore c:a+b:: 2 sin. AB.cos. A; and consequently cos. A = 2 2 CA? + A B’ – CB? Again, in fig. 9, C A+ 2 sin. 2 CA . A B 2 A B’ = CB2 + 2CA, AD; and AD=AB. A-B A + B С ; and, as cos. : sin. the cos. A ; therefore C A' + A B’ = CB2 + 2 CA. 2 2 AB · cos. A; therefore cos. A= С proportion becomes c : a + b :: sin. CA? + A B’ — C B? : COS. or, calling the sides oppo 2 CA. AB' A-B site the angles A, B, and C, respectively a, b, and c, 2 and observing that the cosine of an obtuse angle Formula 6.-In any triangle, as ABC is negative, we have universally cos. A = C A+B b2 +c? -a (fig. 10), c:a-6 :: radius being unity. 2 2 b c' atbtc Formula 8.-In any triangle, as ABC (fig. 10), if be put =S, then cos. A Vrad. 2.S.S 12 + c - od 2 b c For, by the last formula cos. A = and 1 = 2 b c 2 b c A 2 be 68 + c - a? b + c - a? A therefore 1 + cos. A = 2 cos. + ; or, cos. 2 b c S.S be R’S.S-a А dius unity = to radius R; whence cos. be 2 bc rad. ? Sb.S-C Formula 9.-In any triangle, sin. A=V ; adopting the same notation b c 6 с employed in the last formula. A 2 b c 12 + e? a? — (62 + c — 2 b c). ad -For 1 — cos. A= 2 sin. ? 2 2 b c 2 b c 2 b c 2 b c a + b + e a + b + c 6 2 S-6.5 whence sin." to 4b c bc R’S-b.s A rad.'.S b.S radius unity; to radius R, whence sin. 2 bc Examples of the application of the formulæ for the solution of right angled plane triangles. 1. In a plane right angled triangle, as A B C', fig. 11, given AC 480, and ¿ A 53' 8", required AB and BC. AC:cos. A By Prop. VII. AB = and BC= AC -sin. A; these equations solved logarithR R Log. A C, 488, 2.681241 216 Ꭲ Ꭱ I G 0 N 0 Ꮇ Ꭼ Ꭲ Ꭱ Y. 2. Given A B, 1214, and 2 A, 51° 40' 30', to find AC and BC. AB · tan. A AB· sec. A By Prop. VII. we have BC = and AC whence R Log. A B, 1214, 3:084219 Sec. A 51° 40' 30" 10 207523 3.291742 and BC= BC 1535-8 log. 3.186338 A C 1957-7 log. 3. Given A B 63.4, and AC 85:72, to find the other parts. AB· R By Prop. VII. AC · cos. A = AB · R, whence cos. A = AC AB 634 10 tan. A, 42° 18' 4" A Botan. A R 1.802089 B C 57.69 log. 1.761115 4. Given A B 8372:1, and B C 694:73, to find the other parts. BC:R By Proposition VII. A B · tan. A = BC:R, whence tan A= ; and by the saine pro AB Hence A B 8372:1 log. 3.922834 sect. A 4° 44' 37" 10.001490 13.924324 10. AC 8400.9 log. 3.924324 Examples of the application of the formulæ for the solution of oblique angled plane triangles. 1. In a triangle, as ABC, fig. 10, given A B, 376, L A 48° 3', and L B 40° 14', to find the other points. The angle C, being the supplement of the sum of A and B, is 31° 43'. AB · sin. A whence AC = ; and BC = sin. C R? AB · sin. Acosec. C' sin. C AC 242.97 log. 2:385550 BC 279:77 log. 2.446797 2. In a triangle A BC, fig. 12, are given BC 2943, A C 2314, and / B opposite the less of the given sides, = 139024, to find the other parts. 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 B C A' answer to the given data ; and this conclusion agrees also with BC · sin. B that obtained from the formula for finding A, which is sin. A = -; whence the result AC may be the angle A or its supplement; or, in the figure, the result may be the angle A, or its supplement BA'C. Log. BC, 2943 3.468790 The supplement of the sum of B and A is BCA = 86° sin. B 39° 24' 9.802589 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= 13.271379 R? Log. AC, 2314 3364363 BC · sin. BCA' · cosec. BA'C R? 3.468790 3.468790 sin. BCA' 14° 25' 46' 9.396526 cosec. A 53° 49' 46" 10.092984 cosec. BA'C 126° 10' 14" 10.092984 23.561084 22.958300 R? R? BA 3639.8 log. 3.561084 2.958300 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. CB : cos. C From B demit B D a perp. on AC; then CD = prop. VII; and AD= rad. CA-CD. DA · cot. C Then, Formula 2, proposition VIII., CD:DA :: cot. C : cot. A, or cot. A= CD BC · sin. C · cosec. A the angle A B C is the supplement of the sum of A and C; and A B R? tan. 2 2 2 A B 276.286 Log. 2.441359 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 A+B is given, and by the formula referred to AC +CB: AC-CB :: tan. ; 2 A + B B— A _'(AC—CB) tan. A + B B-A A+B B A. 2 and A= and B + 2 2 2. AC + CB Now 180° - 57° 21' 10" + B = 61° 19' 25" = AC + BC = 3826:3 and AC BC = 2 1867.7. Hence. Log. 1867.7 3.271307 A +B 10-262053. Hence A = 19° 24' 35', and B 103° 4' 15', as before. tan. 61° 19' 25' 2 13.533360 Log. 3826:3 3:582779 A-B 2 41° 44' 50'tan, 9.950581 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. By Formula 4, Prop. VIII, we have AC : A B + BC: AB — BC:AD -- DC', whence AD-DC AB + BC · AB BC AC AD-DC and CD= and cos. C= 2 2 AC CD: rad. Now A B + BC = 6577, AB — BC = 939, and 2 AL=8574, whence Log. 6577 3.818028 Log. C D 1423.24 3.153278 10 2 2 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. S ? bic B S and by Form. 9, Prop. VIII., sin. 2 c 1674. ac Promiscuous examples of the application of BC • sin. BCA BA = trigonometrical formulæ to the determination of rad. heights and distances. DC · sin. D · cosect DBC sin. BCA 1. Wanting to determine the height A B of a rad.3 tower (fig. 13), I measured from its base A Ea BC · cos. BCA distance of 230 yards, and then found its 148.305, and AC= angle of elevation B D C = 28° 47', required its rad. • cosect DBC height, allowing five feet for the height of the DC · sin. D · cos. BCA instrument? rad.: DC • tan. BDC 251.94. Here BC = = 126.36 rad. 3. At D (fig. 15), I took BDC the angle yards; whence BA=BC + 5 feet = 128.08 of elevation of a light-house on the top of yards. a hill = 43° 24', and measuring directly towards 2. Being desirous of ascertaining the height it 180 yards, to C, I took again the angle of eleof a steeple B A (fig. 14), at D, I took its vation B C E of its top, 63° 12', and AC E the angle of elevation 16° 4', and measuring directly elevation of its bottom 49° 14', required the towards it, 263 feet to C, I found its angle of height of the tower, and also the height of the elevation BCA= 30° 29', required its height hill on which it stood ? and its distance from C? Here, as in the last example, we have D BC= Here the angle D BC = BCA-BDC= BCE-BDC = 19° 48', and B E DC:sin. D·cosect DBC DC · sin. D cos. DBC · sin. DCE 14° 25'; and BC= rad.? rad." . = or a b 326-03, and by Form. 2, Prop. VIII., EB:EA 7. In a tower besieged there are three steeples, : : tan. BCE : tan. A C E, whence EA = A, B, and C (fig. 19), with the distances of EB · tan. ACE which from each other, the besiegers are actan. BCE quainted from a map of the place; A B is 700, EB · tan. ACE cot. BCE BC 560, and AC 1000 yards. Now the be = 191•41, siegers wishing to plant a battery at D are derad.' and A B=BE- AE = 134.62. sirous of knowing whether from thence the town can be attacked with effect. The angle CDB 4. To determine the distance of a tower on was found to be 12° 15', and the angle B DA the opposite side of a river, (see fig. 16) an 15° 35'; required the distance of D from each officer measured a base A B of 346 yards, and at of the objects A, B, and C. A found the angle CA B 67° 14', and at B the On A C let a segment of a circle be described angle C BA = 48° 28', what are the distances of °C, the tower, from A and B, the extremities ADB; and on В A imagine the segment of a to contain the sum of the angles CD B and of the measured base? circle to be described, containing the angle The angle A C B being the supplement of the ' AD B, the intersection of these circles D, as is sum of Å and B is 64° 18'. Hence AC= obvious from the principles of geometry, will be AB · sin. B · cosect C = 287-4, and BC= the position of the battery. rad.? Join B D and let it meet the circumference in AB · sin. A · cosect C E, then the angle C EA, which is the supple= 334.0 rad.? ment of C DA, is 152° 10', and the angle C A E 5. To find the distance between a house and the angle CDE = 12° 15', and ECA= a mill, neither of which I could get near, I EDA = 15° 35'. Hence C E = measured a base A B (fig. 17), = 820 CA. sin. CAE: cosect. CEA =454:438; and yards; at A I observed the angle DAB = 80° rad.? 28', and CAB = 40° 43'; and at B I observed calling BC, a, A C, b, and AB, c; and putting the angle C BA= 100° 12', and D BA = 52° a + b + c S for 46', required the distance C D? -, we have, formula 8, proposition 2 The angle A D B being the supplement of the С sum of DAB and ABD is 46° 46'; and the VIII., cos. rad. S.S. C angle A C B being the supplement of the sum of 2 CAB and CBA is 39° 5°; also the angle D AC, 21° 19' 53"; whence A CB = 42° 39' 46", and which is the difference of D A B and CA B, or consequently BCE=AC B-EC A= 27° 4 39° 5'. 46". Then in the triangle BCE we have BC AB:sin. ABD. cosect ADB 560, CE=454:438, and the angle BCE= Now AD= rad. 2 27° 4' 46", to find the angle C B E, which, by 896.093; and AC operations already sufficiently exemplified, is AB · sin. BAC · cosect ACB found = 50° 50'54". Now in the triangle C BD, = 848:444; wehave given the angle CDB= 12° 15' and the rad. 2 ADC +ACD 180° - CAD angle CBE = 50° 50' 54”, and consequently and = 70° the angle BCD= 116° 54' 6"; and the side 2 2 BC being given, we have B D = 27' 30'; and by Form. 3, Prop. VIII., tan. BC · sin. BCD · cosect. CDB ADC ACD = 2353.7 and rad." 2 CB · sin. CBD · cos. CDB CD= rad.? 2061.6. wbence ADC=66° 3' 29", and ACD=740 ON SPHERICAL TRIGONOMETRY. 51' 31" ; and CD= AD · sin. DAC · cosect ACD Prop. I.-In any right angled spherical tri = 585.26. rad.? angle, the rectangle of radius, and the sine of 6. To determine the distance A B (fig. 18), equal to the rectangle of the tangent of the other either of the sides containing the right angle, is between the extreme points of the entrance side, and the cotangent of the angle opposite to of a barbour, the distance of each from that side. point C inland, and the angle A C B were measured ; AC = 896 yards, BC= 1014, and the angle, right angled at "B; and let D be the Let ABC (fig. 20), be a spherical triangle C 28° 44', required the distance A B? centre of the sphere. From B, in the plane A +B А. В Here = 75° 38'; tan. BAD, draw BE perpendicular to DA, and B from E in the plane A C D draw E F also perA+B AC – BC · tan. pendicular to D A, meeting D C produced in F, and join F B. Then as A D is perpendicular to 2 =13° 33' 32', whence E F and E B at their point of meeting, it is perAC + BC pendicular to the plane E B F, and therefore the A= 89° 11' 32", and B= 62° 4' 28"; and A B plane BD A, which passes through AD, is also AC · sin. C cosect B = 488.022. perpendicular to the plane B EĚ. Now, as the rad.? angle A B C is a right angle, the plane BDC is a 2 |