The side AB is not found instrumentally for a reason similar to that before given. 2. The perpendicular AC=200, and the base BC=110 of a right angled triangle ABC being given, required the hypothenuse AB, and the angles A and B? Ans. AB=228.254, ang. A=28° 49, ang. B=61° 11. 3. Given AC=4, and BC=3, to find AB, and the angles A and B. Ans. AB=5, ang. A=36° 52′, ang. B=53o 8′. 78. Case 3. The hypothenuse AB and the angle B being given, to find the sides AC, CB, and the angle A. First, since the angle at B is given, the angle A=90°-B. Secondly, (Art. 63.) AB: AC:: radius sin B.AB : sin ang. BAC= radius ; and log. AC =log. sin B+log. AB-log. radíus; whence AC is found both by natural numbers and logarithms. 2 Thirdly, ABAC)2+CB (47. 1.) .. CB-AB-AC and CB= √AB+AC.AB—AC (cor. 5.2.); also log. log. AB+AC+log. AB-AC CB 2 natural numbers and logarithms. B CB is found, both by EXAMPLES. 1. Given the hypothenuse AB=165, and the angle B 35° 30', to find the sides AC, CB, and the angle A. By calculation. First, ang. A=90°—B= (90°—35° 30′=) 54° 30′. sin B.AB Secondly, (by natural numbers) AC rad (since rad=1, sin 35° 30′ x AB=) .580703 × 165=95.815995; but the same may be done more readily by logarithms; thus, because log. AC=log. sin B+log. AB-log. rad. To log. sin B. or 35° 30′= Add log. AB. or 165: And from their sum= Remains log. AC 95.816= Thirdly, CB=AB+ AC.AB-AC= {√165+95.816 × 165–95.816=/260.816 × 69.184⇒ 18044.294144) 134.329. The same by logarithms, log. CB= log.`AB+AC+log. AB—AC 1. Extend from (radius or) 90° to 35° 30′ (=ang. B) on the line of sines; this extent will reach from 165 (backwards) to about 95% on the line of numbers, for the side AC (opposite the ang. B.) 2. Extend on the line of sines, from 90° to 54° 30' (comp. B.); this extent will reach on the lines of numbers from 165 to about 134 for the side CB. Ex.-2. Given the hypothenuse AB=25, and the angle B=49°, to find the sides AC, CB, and the angle A? Ans. AC 18.893, CB=16.4017, ang. A=41°. 3. Given AB=100, and the angle A=45°, to find the rest? Ans. BC=AC=70.7108, ang. B=45°. 79. Case 4. One side AC, and its adjacent angle A being given, to find the other sides AB, BC, and the remaining angle B. B AB=log. AC+log. rad. —log. sin B. EXAMPLES.-1. Given the perpendicular AC=1023, and the angle A=12° 45′; to find the angle B, and the remaining sides AB, BC. By calculation. First, ang. B=90°-A=(90°-12° 45') 77° 15'. Secondly, CB= sin A.AC .2206974 x 1023 -=231.4812; .9753423 and by logarithms, log. CB=log. sin A+log. AC-log. sin B; Subtract log. sin, B 77° 15′= .... 9.9891571 Gives log. CB. 231.4812—– ..... 2.3645158 AC.rad. 1023 × 1 Thirdly, AB= =1048.862. sin B .9753423 And by logarithms, log. AB=log. AC+log. rad-log. sin B; that is, to log. AC 1023== Add log. radius= And from the sum= ..... 3.0098756 10.0000000 ... 13.0098756 1. To find CB, extend from (sin B, to sin A, that is, from) sin 770 to sin 120; this extent will reach on the line of num.. bers from (AC) 1023 to 2314. 2. To find AB, extend from (sin B to radius, that is, from) 77° to 90° on the sines; this extent will reach from 1023 to about 1049 on the numbers. Ex.-2. Given the perpendicular AC=400, and the angle A=47° 30′, to find the hypothenuse AB, the base BC, and the angle B? Ans. AB=592.072, BC=436.52, ang. B= 42° 30'. 3. Given AC=82, ang. A=33° 13', to find the rest? Ans. AB=97.9, CB=53.69, ang. B=26° 47′. SOLUTION OF THE CASES OF OBLIQUE ANGLED TRIANGLES. The foregoing calculations are effected both by natural numbers and logarithms, serving as a useful exercise for the learner; but principally to shew, that both methods terminate in the same result. Trigonometrical operations are however seldom performed by the natural numbers, and therefore, in the following cases, we shall employ only the logarithmic process. SO. Case. 1. Let there be given the two angles B and C, and the side AC opposite to one of them; to find the angle A, and the sides AB and BC. First, the angles B and C being given, and A=180o— B+C, the angle A will be known. Secondly, (Art. 67.) AC : AB: sin B sin C. AB= AC. sin C sin B : ; or by logarithms, log. B A C AB=log. AC+log. sin C-log. sin B ; ·.: AB is known. Thirdly, (Art. 67.) AC : CB :: sin B: sin A ·: CB= AC. sin A sin B By logarithms, log. CB=log. AC+log. sin 4—log. sin B; CB is known. EXAMPLES.-1. Given the angle B=46°, the angle C=59°, and the side AC (opposite B)==120; to find the angle A and the sides AB, BC. By construction. From any scale of equal parts take AC120, at C make the angle ACB=59, and at A make the angle CAB (180°B+C=180°—46° +59°=) 75°; then take the length of AB, and of BC respectively in the compasses, and apply them to the above-mentioned scale, and AB will=143, BC=161. By computation. 1. Log. AB log. AC+log. sin C-log. sin B 2. Log. CB=log. AC+log. sin A-log. sin B. 1. Extend on the sines from 46' (ang. B), to 59° (ang. C); this extent will reach on the numbers from 120 (AC), to about 143 (AB). 2. Extend from 46° to 75o on the sines; this extent will reach from 120 (AC), to about 161 (CB), on the numbers. Ex. 2. Given the angle A=58° 43′, the angle C=74° 7′, and the side AB=610; to find the angle B, and the sides AC, CB? Ans. ang. B=47° 10′, AC=465.08, CB=542. 3. Given the side AB=1075, the angle B=34° 46′, and the angle C=22° 5′; to find the rest? Ans. BC=2394, AC=1630.5, ang. A=123° 9'. 81. Case 2. Let there be given the two sides AB, AC, and the angle B, opposite AC; to find the angle BAC and C, and the remaining side BC. • This case will be always ambiguous when the given angle B is acute, and AB greater than AC, (as in the first example); for the above expression is the sine of both AxB= Azx, or of its supplement AzB (for the sine of an angle and the sine of its supplement are the same, by cor. Art. 12.); consequently the angle A will be either BA or BAz, according as the angle AxB, or its supplement AzB be taken; and the corresponding value of BC will be either Br or Bz. But if the given angle be either obtuse, or a right |