SOLUTION OT TRIANGLES.] 



NAVIGATION. 



1039 



And first as to the table of natural sines, cosines, (to. 

 By referring to the diagram at p. 1037, we see that this 

 table supplies, already computed for us, the numerical 

 lengths of the sides of two similar right-angled triangles 

 O Bn, O<A, whatever be the angle O, from O = 0", up to 

 O = 90 1 . We may be quite sure, therefore, whatever be 

 the magnitude or shape of any right-angled triangle met 

 with in practice, that two triangles, equiangular to it, 

 will always be found in the table ; the sides of the one, 

 O Bn, being computed to the scale O B, the hypotenuse, 

 = 1, and the sides of the other to the same scale, the 

 base O A, = 1. The table thus f urnishes us with the 

 perpendicular and base of every possible right-angled 

 triangle whose hypotenuse is 1, and also with the per- 

 pendicular and hypotenuse of every possible right- 

 angled triangle whose base is 1. 



Now the sides about the equal angles of equiangular 

 triangles are proportional (Euc., Prop. 4, Book VI.), so 

 that if one side only of a triangle be known, and all the 

 sides of another, equiangular to it, be given, we can com- 

 pute the unknown sides of the former by simple propor- 

 tion. It follows, therefore, that if one side only, of 

 any right-angled triangle, with which we may be en- 

 gaged in practice, be measured, we shall be able to cal- 

 culate each of the other sides, in this manner, provided 

 only we know where to look for the triangle, equiangular 

 to the proposed triangle, in the table. 



As already observed, there are always two such equi- 

 angular triangles in every complete table of natural 

 <, cosines, <tc., so that we have a choice as to which 

 shall be compared with the triangle to be calculated ; 

 ve must, of course take which ever we may be 

 careful to compare base with base, perpendicular with 

 perpendicular, and hypotenuse with hypotenuse. The 

 sides of the two triangles O Bn, OtA, computed in the 

 table, do not, however, go by these names : the base of 

 the triangle O Bn is called cosini of the angle O, the 

 perpendicular is called tine of the angle O, and the hypo- 

 tenuse is called radius, or 1. Also the base of the similar 

 triangle O<A is radius or 1, the perpendicular is tangent 

 of the angle O, and the hypotenuse is secant of the angle 

 O : the angle O, expressed in degrees and minutes, is thus 

 a sufficient guide to direct us to that part of the table where 

 the equiangular triangle we are in search of is to be found. 

 It is always best to fix upon that one of the two tabu- 

 lar triangles in which the unit-length, or radius, corre- 

 sponds to the given side of the triangle proposed for 

 calculation. Thus, if the given side be the hypotenuse, 

 we should compare our triangle with the tabular triangle 

 O Bn, because in this the hypotenuse is 1 ; wo should 

 then have to compare the perpendicular of the proposed 

 triangle with Bn, that is, with sin. O, and the base with 

 On, that is, with cos. O. For example, if the angle equi- 

 valent to O, and the hypotenuse were given to find the 

 ndicular and the base, then, referring to the neces- 

 sary particulars, on the page of the table headed by the 

 degrees and minutes in the given angle, we should be 

 supplied with the first three terms of the proportion. 



1 : sin. O : : given hyp. : required perp., 

 Consequently the required perp. = yp. X 



- hyp. X sin. O. 



Also, 1 : cos. O : : given hyp. : required base, 



. , hyp. X cos. O 

 Consequently the required base ^-5- 



hyp. X cos. O. 



As the first term of each proportion is 1, all division is 

 thin avoided; and the operation reduced merely to the 

 multiplication together of two factors. 



On the other hand, if the base were given to find the 

 perpendicular and hypotenuse, the other tabular triangle, 

 O<A, would be the better to compare with the proposed 

 one, because the radius or unit-line here is O A, the base, 

 corresponding to the side given, and the proportions would 

 be 1 : tan. O : : given base : required perp. , 



and 1 : sec. O : : given base : required hyp., 



.'. required perp. = base X tan. O, and required hyp. 



base X sec. O. 



As to the tables of logarithmic sines, cosines, (fee. , little 

 need be said iu addition to what has already been stated ; 

 they furnish the logarithms of the tabular numbers in 

 the table of natural sines, cosines, <kc. , with the addition 

 of 10 to each logarithm, as before explained. By using 

 the logarithmic tables, we convert the multiplication of 

 the two factors adverted to above, into the addition 

 of the corresponding logarithms. Sufficient practical 

 details of these operations will be given in the next article. 



It may be noticed here, however, that in what is said 

 above, as to the assistance afforded by trigonometrical 

 tables, our observations have had exclusive reference to 

 right-angled triangles ; we shall shortly see that they are 

 equally available for oblique-angled triangles; but, it will 

 strike the learner as a remarkable fact, that, notwith- 

 standing the complicated character of the curve, traced 

 on the surface of the globe by the course of a ship, all 

 the calculations in reference to the different sailings, 

 involve only right-angled triangles the simplest kind of 

 triangles with which trigonometry has to deal ; we shall, 

 therefore, give a distinct article on the calculation of the 

 sides and angles of right-angled triangles. 



To CALCULATE THE SIDES AND ANGLES OP RIOHT- 

 ANOLED TRIANGLES. The sides and angles of any tri- 

 angle make up what are called the six parts of the 

 triangle ; and if any three of these six parts, provided 

 they are not the three angles, be given, the remaining 

 three can be found by calculation. In a right-angled 

 triangle, one ftf the six parts is always known namely, 

 the right angle ; so that a side and one of the acute 

 angles, or two sides, being measured, we can always cal- 

 culate the remaining parts. 



The reason that the three angles alone will not suffice 

 for the determination of the other three parts the three 

 sides is obvious from the simplest principles of geome- 

 try ; for all equiangular triangles are alike, as respects 

 the equality of the angles ; so that, from the angles alone, 

 we could not conclude to which one, out of an infinite 

 variety of similar triangles, the proposed angles are con- 

 sidered to belong, for they belong, in reality, equally to all. 



In a right-angled triangle the given parts must there- 

 fore be either : 



1. A side and one of the acute angles, or, 



2. Two of the sides, to find the remaining parts. 



I. When the Hypotenuse and one of tlie Acute Angles are 



yicen. 



Let the hypotenuse AB (Fig. 2), and the angle A bo 

 given ; then turning to the diagram at page 1037, we 



Fig. 2. 



should compare the given 

 side A B, with the radius 

 O B, as directed above ; so 

 that OBn would be the 

 tabular triangle equian- 

 gular to the triangle ABC. 

 And, therefore, if c denote 

 the numerical length of 

 AB, since the numerical 

 length of O B is 1, A B 

 = O B X c ; consequently, 

 a and 6 being the numerical lengths of B C, and A C, we 

 have, since the like sides are proportional, 



B C = B n X c, and A C = O n X c, that is, 



o = c sin. A, and 6 = c cos. A. 

 Hence the following rule : 



ROLE. Multiply the given hypotenuse by the sine of 

 Fig. s. the angle at the base ; the pro- 



duct is the perpendicular. 

 Multiply the given hypotenuse 

 by the cosine of the angle at 

 the base : the product is the 

 base. 



Or work by the following 

 formula, 



perp. = hyp. X sin., ang. 

 at base ; base = hyp. X cos., 



ang. at base, 



which by logarithms will be, 

 log. perp. = log. hyp. + log. sin. , ang. at base 10 ; 

 log. base = log. hyp. -j- log. cos. , ang. at base 10. 



