TRIGONOMETRY. 



angle ABC, leaves the angle E B C. But if the perpen- 

 dicular (hall fall without the triangle, the angle ABC 

 (hould have been fubtrafted from ABE. Since as the 

 perpendicular B E is taken for one of the lateral parts, the 

 middle part in the triangle A B E is the angle B, and the 

 conjoint part A B ; in the triangle E B C the middle part is 

 the angle B, and the conjunft part B C ; the cotangent of 

 the fide B C is found by fubtrafting the cofine of E B A 

 fi-om the fum of the cotangent of A B, and the cofine of 

 E B C. The example of the preceding cafe is eafily applied 

 to this. 



5. Given two fides A B, 66° 45', and B C, 39° 29', 

 with the angle A oppofite to one of them, 43° 20' ; to 

 find the third fide A C. 



Letting fall, as before, the perpendicular BE; in the 

 right-angled triangle ABE, from the given angle and hypo- 

 thenufe A B, find the fide A E. Since affuming BE for a 

 lateral part in the triangle A E B, A B is the middle part, 

 and A E is the feparate part ; and in the triangle B E C, 

 B C is the mean part, and E C a disjundl part ; the co- 

 fine of E C is found by fubtrafting the cofine of A B from 

 the fum of the cofines of A E and C B. If then the feg- 

 mcnts A E and E C be added together, or in cafe the per- 

 pendicular fall without the triangle, be fubtrafted from each 

 other, the fide A C will be had. 



6. Given two fides AC, 65° 31', and A B, 66° 45', to- 

 gether with the included angle A, 43° 20'; to find the third 

 fide B C oppofite thereto. 



Letting fall the perpendicular B E, find, in the right-angled 

 triangle, the fegment A E ; which, fubtrafted from A C, 

 'leaves E C. If the perpendicular fall without the triangle, 

 AC is to be fubtrafted from A E. Since by affuming the 

 perpendicular B E for a lateral part in the triangle A E B, 

 A B becomes a middle part, and A E a feparate part ; in 

 the triangle E B C, C B is the middle part, E C a feparate 

 part : the cofine of B C is found by fubtrafting the cofine 

 of A E from the fum of the cofines of A B and E C. 



7. Given two angles A, 43° 20', and B, 79° 3', together 

 with the fide C B, 39° 29', oppofite to one of them ; to find 

 •the fide A B adjacent to both. 



Letting fall the perpendicular C D from the unknown 

 angle C, to the oppofite fide A B ; and that falling within 

 the triangle ; from the given angle B, and the hypothenufe 

 B C, feek in the right-angled triangle B C D for the fegment 

 B D. Since affuming the perpendicular C D for a lateral 

 part in the triangle C D B, D B is the mean part, and the 

 angle B a conjundl part ; and in the triangle C D A, A D 

 is the middle part, and the angle A a conjunft part : the fine 

 of the fegment A D is found by fubtrafting the cotangent 

 of the angle B from the fum of the fine of D B, and the co- 

 tangent of the angle A. If then the fegments AD and 

 D B be added, or in cafe the perpendicular fall without the 

 triangle, be fubtrafted front each other, the refult will be the 

 fide A B required. 



8. Given two fides A B, 66° 45', and B C, 39° 29', with 

 the included angle 79° 3' ; to find the angle A oppofite to 

 one of them. 



Letting fall the perpendicular C D, find the fegment B D, 

 as in the preceding problem. This, fubtrafted from A B, 

 leaves A D. If the perpendicular fall without the triangle, 

 A B is to be added to D B. And fince by affuming the 

 perpendicular CD for a lateral part in the triangle C D B, 

 B D is the middle part, and the angle B a conjoint part ; 

 and in the triangle C D A, AD is the middle part, and 

 the angle A a conjoint part, the cotangent of the angle A is 

 formed by fubtrafting the fine of D B from tlie fum of the 

 cotangent of the angle B, and of the fine of A D. 



9. Given two angles A, 43° 20', and B, 79" 3', togethef 

 with the adjacent fide A B, 66° 45' ; to find the angle C 

 oppofite to the fame. 



From one of the given angles B, letting fall the perpen- 

 dicular B E to the oppofite fide AC; in the right-angled 

 triangle ABE, from the given angle A, and hypothenufe 

 A B, wc find the angle ABE; which, fubtrafted from 

 ABC, leaves the angle E B C. In cafe the perpendicular 

 fall without the triangle, ABC is to be fubtrafted from 

 ABE. Since by affuming B E for a lateral part in the 

 triangle C E B, the angle C is a middle part, and the angle 

 C B E a disjunft part ; and in the triangle ABE, the angle 

 A is the middle part, and the angle ABE the disjoint part : 

 the cofine of the angle C is found by fubtrafting the fine 

 of the angle ABE, from the fum of the cofine of the angle 

 A, and the fine of E B C. 



10. Given two angles A, 43° 20', and C, 82° 34', together 

 with the fide B A, 66° 45', oppofite to one of them ; to find 

 the other angle. 



From the angle B fought, let fall a perpendicular B E ; 

 and in the right-angled triangle A E B, from the given 

 angle A, and hypothenufe B A, 'find the angle A B E. 

 Since affuming the perpendicular E B for a lateral part in 

 the triangle E C B, the angle C is the middle part, and the 

 angle C B E a disjunft part ; and in the triangle ABE, the 

 angle A is the middle part, and the angle A B E is a dif- 

 junft part : the fine of the angle E B C is found by fub- 

 trafting the cofine of A from the fum of the cofine of C, 

 and of the fine of A B E. If then A B E and E B C be 

 added, or in cafe the perpendicular fall without the triangle, 

 be fubtrafted form each other, the refult will be the angle 

 required ABC. 



1 1 . Given the three fides, to find an angle oppofite to one of 

 them. I. If one fide A C (P/fl/f II. ^_?. 8.) be a quadrant, 

 and the leg A B be lefs than a quadrant, find the angle A. 

 Continue A B to F, till A F become equal to a quadrant ; 

 and from the pole A draw the arc C F, to cut the arc B F 

 at right angles in F. Since in the right-angled triangle C B F, 

 we have given the hypothenufe B C, and the fide F B, or 

 its complement A B to a quadrant ; we iliall find the per- 

 pendicular C F, which being the meafure of the angle CAB, 

 that angle is found of courfe. 



2. If one fide A C be a quadrant, and the other A B 

 greater than a quadrant, feek again the angle A ; from A B 

 fubtraft the quadrant A D, and from the pole A defcribe 

 the arc C D, cutting the arc A B at right angles in D. 

 Since in the right-angled triangle C D B, the hypothenufe B C, 

 and fide D B, or excefs of the fide A B beyond a quadrant, 

 is given, the perpendicular C D will be found iis before, 

 which is the meafnre of the angle A required. 



3. If the triangle be ifofceles, and A C ^ C F ; and the 

 angle A C F be required ; bifeft A F in D, and through 

 D and C draw the arc D C. Since D C is perpendicular 

 to A F, the angles A and F, and A C D and D C F, are 

 equal ; from the hypothenufe A C, and leg AD, given in 

 the right-angled triangle A C D, we find the angle A C D ; 

 the double of which is the angle required A C F : and from 

 the fame data may the angle A or F be found. 



4. If the triangle be fcalenous, and the angle A {Plate 111. 

 Jig. 5. ) be required ; from C let fall the perpendicular C D, 



and feek the half-diifcrence of the fegments A D and D B, by 

 faying, as the tangent of half the baie A B is to the tangent 

 of half the fum of the legs A C and C B ; fo is the tangent 

 of their half-difference to the tangent of the half-difference of 

 the fegments A D and D B : add then the half-difference 

 of the fegments to the half-bafe, to find the greater feg- 

 ment ; and fubtraft the fame from the fame for the lets- 

 7 Thus 



