SOLUTION OP TRIANGLES.] 



NAVIGATION. 



1041 



log. AB = log. AC + log. sec. A 10. 

 10 



AC = 327 2.5145 



sec. A, 54 17' 10-2338 



AB, 500 27483 



The logarithmic operation might have been performed 

 as readily by using the formula for A B, as before, 

 which gives 



log. A B = log. A C log. cos. A + 10. 

 The work by this formula will be as follows : 



10 



A C = 327 2 5145 



cos. A, 54 17' -9-7CU2 



AB=5CO 27483 



2. In the triangle ABC (Fig. 3), are given A C = 195, 

 and the angle A = 47 55', to find B C and A B. 



Ana. B C = 210, A B = 291. 



3. Given A C = 288, and the angle A = 53 8', to find 

 B C and A B. Ans. B C = 384. A B = 480. 



4. Given A C = 421, and the angle A = 36 J 6', to find 

 BCandAB. Ans. BC = 307, AB = 521. 



5. Given A C = 625, and the angle B = 41 15', to find 

 BCandAB. Ans. B C = 713, A B = 948. 



6. Given the base of a right-angled triangle 346 i, and 

 the opposite angle 54 36' : required the perpendicular 

 and hypotenuse. Ans. Perp. 240 '2, Hyp. 425-1. 



III. When the hypotenuse and one of the other sides are 



given. 



Let the hypotenuse A B (Fig. 3), and the perpendicular 

 B C be given ; then comparing A B with the radius or 

 unit line O B in the diagram at page 1037 (Fig. 1), the 

 tabular triangle O B will be that to which the proposed 

 triangle is to be compared : and since AB is c times O B, 

 therefore B C is c times Bn ; that is to say, 



a = B C = c sin. A, therefore sin. A 



If the base were given instead of the perpendicular, then 

 we should have 



6 = A C = c cos. A, therefore cos. A = - 



The rule is therefore as follows : 



RULE. Divide the perpendicular by the hypotenuse : 

 the quotient will be the sine of the angle at the base. 

 Divide the base by the hypotenuse : the quotient will be 

 the cosine of the angle at the base. 



An angle being thus determined, tlie remaining side of 

 the triangle may be found by either of the preceding rules. 

 Or, without first finding an an<'le, the third side of any 

 right-angled triangle may be determined from the other 

 two, by the 47th of Book I. of Eur.lid ; for since c 2 = a 2 

 -f-b 2 ; .'. a 2 = c 2 ft 3 , and V = c 2 - a 2 ; the working for- 

 mulae for finding the angles, when the hypotenuse and a 

 side are given, are 



perp. 1 1:1*0 



sin. ang. at base = - : cos. ang. at base = r 



BTpT hyp. 



Or, by logarithms, 



log. sin. ang. at base = log. perp. log. hyp. -(-10 ; 

 log. cos. ang. at base = log. base log. hyp. +10. 



EXAMPLES. 



1. In the right-angled triangle ABC (Fig. 3), are given 

 the hypotenuse A B = 480, and the perpendicular B C = 

 384, to find the angles A, B, and the base A C. 



1. To find the angles A, B. 

 Without logarithms. 

 BC 



'IB' 



< sin. 53 8' 



sin. A < 



480)384(-8' 



384 



With logarithms. 



log. sin. A = log. BC-log. AB + 10 

 10 



B 0,384 2-5843 



AB, 480 -2-6812 



Bin. A, 53 8' 



9-9031 



Consequently A = 53 8', .'. B = 90 -53 8' =36 



2. To find ihe base A C. 

 A C = A B cos. A 

 cos. A, 53 8', = -6 

 .'. 480 X "6 = 283 = A C 



AC is thus determined with very little trouble, 

 log. AC = log. AB + log. cos. A- 10 

 -10 



AB, 480 2-6812 



cos. A, 53 8' . . . . 9-7781 



A C, 288 2-4593 



To find the base without the aid of the angle, we have 

 AB 2 -BC 2 = (AB + BC)(AB-BC) 

 96 = 82944 



82944 (288 ; or by logs., log. 864 = 2-9365 

 4 log. 96 = 1-9823 



48)429 

 384 



568)4544 

 4544 



2)4-9188 

 log. 288 = 2-4594 



Consequently A C = 288. The sum of the two logs, on 

 the right is divided by 2, because the half of the log. of a 

 number is the log. of the square root of that number. 



2. Given the hypotenuse A B = 237, and the perpendi- 

 cular B C = 197, to find the angles and the base. 



Ans. A _ 56" 13', B = 33 47', Base = 132. 



3. Given the hypotenuse A B = 645, and the perpendi- 

 cular B C = 407 "4 to find the angles and the base. 



Ans. A = 39 KX, B = 50 50', Base = 500. 



3. Given the hypotenuse A B = 400, and the base A C 

 = 236, to find the angles and the perpendicular. 



Ans. A = o3 51', B = 30 &, Perp. = 323. 



4. Given the hypotenuse AB = 54 '68, and the base 

 A C = 35 '5, to find the angles and the perpendicular. 



Ans. = 49 31', B = 40 29', Perp. = 41 -6. 



6. Given the hypotenuse of a right-angled triangle = 

 779 '8, and the perpendicular = 725, to find the vertical 

 angle and the base. Ans. 21 30', and 287.1. 



6. Given the hypotenuse = 780 "6, and the base = 

 607 '6, to find the angle at the base, and the perpendi- 

 cular. Ans. 38 53', and 490. 



IV. IFhcn the base and perpendicular are given. 



The base A C = b, and the perpendicular B C = a being 

 given, we are to compare the triangle with the tabular 

 triangle O A.t, of which the base O A = 1 : we thus have 

 B C = 6 times At, that is 



a = 6 tan. A .'.tan. A =-T- 







ConsequenMy the rule for finding the angle A at the base 

 is as follows : 



RULE. Divide the perpendicular by the base : the 

 quotient will be the tangent of the angle at the base. Or, 

 divide the base by the perpendicular : the quotient will 

 be the tangent of the angle at the vertex. The hypo- 

 tenuse may be found without first finding the angles by 

 Euclid, 47th of Book I. : thus c 2 = a 2 + 6 2 . The formula 

 for finding either angle is therefore 



tan. ang at base 



perp. 

 - 



TOL. I. 



. 

 ; tan. ang. at vertex 



6 a 



base 



