numbers, AB2=300-2202-90000-48400=41600, and extracting the square root, AB=204=the base.

Example 2. Art. 142. AC+BC=135+108=243=the sum of the sides, whose logarithm=2.3856063. And AC -BC=135–108=27, the difference of the sides, whose logarithm=1.4313638. The sum of 2.3856063, and 1.4313638 is 3.8169701, which divided by 2, becomes 1.9084850, the natural number belonging to which is 81, the length of the base.

SOLUTIONS OF OBLIQUE ANGLED TRIANGLES.

The letters a. c. as connected with one of the terms of a proportion, are an abbreviation for Arithmetical Complement. In this section, where they are prefixed, they denote 10 diminished by the logarithm of the first term; and in this case, the sum of the first three terms diminished by 10=the loga. rithm of the fourth term. If a the 1st term, z=the 4th, and if m+n—a=z, it is plain See Art. 53 seq.

and m, n=the 2d and 3d; that m+n+10—a—10—z.

CASE I. Given the two angles and a side, to find the remaining angle and the other two sides.

Example 2. Figure 27. By THEOREM I. To find the side a sin Bb::sin Aa; and, by the tables, a. C. 0.1487535 1.8512583:19.9803639 1.9803757

95.58. Again, to find the side c, sin Bb::sin Cc; and, by the tables, 0.1487535: 1.8512583::9.6668238: 1.6668356= 46.43=45° 14'. Therefore the side a=95.58, and the side c=46.43. The angle B=180°-(107° 6′+27° 40′)= 45° 14'.

CASE II. Given the two sides and an opposite angle, to find the remaining side and the other two angles.

Example 1. Figure 28. By THEOREM I. To find the angle B, asin A::b:sin B: and, by the tables, a. c. 8.3010300 9.7621775:: 1.8450980: 9.9083055=54° 3′ 50" the acute angle B. The obtuse angle=180°-(54°3′ 50"+125° 56′10′′)=18° 43′50". To find the side c, sin A; a sin Cc; and by the tables, a. c. 0.2379225: 1.6989700

Therefore the acute

angle B 54° 3′ 50′′, and the side c=27.76.

To find the side AB', sin A ; CB': :sin C ; AB'; and by the tables, a. c. 0.2378225: 1.6989700 :: 9.9999758 : 1.9367683-86.45-AB'.

Example 2. Figure 29. By THEOREM I. To find the angle B, a sin A::b: sin B; and, by the tables, a. c. 8.14266759.9521055::1.8061800 9.9009530=52° 45'

25". To find the angle C, add 52° 45′ 25′′, the angle B, to 63° 35', the angle A, and subtract the sum from 180°; the remainder is 63° 49′ 35′′, the angle C. To find the side c, sin A asin C: c; and, by the tables, a. c. 0.0478945 : 1.8573325:9.9523875 1.8576145=72.05. Therefore, 52° 45′ 25′′ the angle B, 63° 49′ 35′′ the angle C, and 72.05 the side c.

To

Example 3. Figure 29. By THEOREM I. To find the angle B, a sin A::b: sin B; and, by the tables, 1.8325089 : 9.9299891 :: 1.6720979: 9.7695781-36° 2′ 4′′. find the angle C, 36° 2′ 4′′+121° 40′, the angle A=1° 15° 42' 4", which subtracted from 180° leaves 22° 17′ 56′′, the angle B. To find the side c, sin A a::sin Cc: and, by the tables, 9.9299891: 1.8325089::9.5791616; 1.4816814 30.3 the side c nearly. Therefore 36° 2′ 4′′ the angle B, 22° 17′ 56′′=the angle C, and 30.3=the side c.

CASE III. Given two sides and the included angle, to find the remaining side and the other two angles.

Example 1. Figure 30. By THEOREM II. To find the angles B, C, (b+c) (b−c)::tan (B+C): tan (BC). Substitnting numbers, 92:14::tan 76° 53′ tan BC: and by the tables, 1.9637878: 1.1461280::10.6326181 : 9.8149583=33° 8′ 50′′ diff. between the angles B, C ; therefore (Alg. 341,) the greater angle C=110° 1′ 50′′ and the less=53° 44' 10". To find the side a, sin Bbsin A:a; and by the tables, 9.8396948 : 1.5910646:: 9.6454496: 1.3968194-24.94. Therefore 43° 8′ 50′′ the angle B, 110° 1' 50" the angle C, and 24.94 the side a.

Example 2. By THEOREM II. To find the angles B and C, (b+c): (b−c): :tan B+C: BC. Substituting num

bers, 185 33::tan 39° 15': tan B C ; and, by the tables, 2.2671717: 1.5185139 :: 9.9122403: 9.1635825-8° 17′ 24" one half the difference between the angles B and C ; and 39° 15' one half their sum: Therefore 47° 32′ 24′′: the angle C, the greater angle, and 30° 57′ 36′′ the angle B, the less. The angle C is the greater angle, because it is oppposite c, the greater side. (Art. 146.) To find the side a, sin B: b::sin A: a, and by the tables, a. c. 0.2886867 : 1.8808136::9.9911927: 2.1606930-144.8=the side a. Therefere the angle B-30° 57' 36", the angle C-47° 32′ 24", and the side a=144.8.

CASE IV. Given the three sides, to find the angles.

Example 1. Figure 31. By THEOREM III. To find the angle A, making the hypothenuse of the right angled tri angle ABC, radius, AC: R::AP: cos A; and, by the tables, 1.5440680: 10.0000000::1.4126285): 9.8685605=42° 21′ 57′′. To find the angle B, BC: R::BP : cos B: and, by the tables, 1.4313638: 10.0000000 :: 1.1185954 : 9.6872316-60° 52′ 42′′. The sum of the angles A and B=103° 14′ 39′′, which subtracted from 180°, leaves 76° 45' 21", the angle A. Therefore the angles A, B and C, are 42° 21′ 57′′, 60° 52′ 42′′, and 76° 45′ 21′′.

104, whose log. 2.0170333
174, whose log.=2.2405492
18, whose log.1.2552725

Example 2. Figure 31. By THEOREM III. By the conditions of the question, AB, the longest side AC+BC, the other two AC-BC. the difference By THEOREM III, (Art. 145,) 2.0170333: 2.2405492 :: 1.2552725: 1.4787884-30.11538=the difference of the segments made by the perpendicular. As the sum of the segments, or the longest side of the triangle=104, and the diff. of the segments 30.11538, by Alg. 341, 67.05769= the greater segment, and 36.94231=the less. To find (Art. 135,) the angle A, making AC radius, AC: R::AP cos A. Substituting numbers, 96: 1:67.05769 : cos A;

and, by the tables, 1.9822712 10.0000000 :: 1.82640377543: 9.84413257543-45° 41′ 48′′the angle A. To find the angle B, making BC radius, BC: RBP : cos B. Substituting numbers, 78: 1::36.94 : cos B; and,

by the tables, 1.8920946: 10.0000000 :: 1.5674969: 9.6754023=61° 43′ 27′′ the angle B. To find the remaining angle C, subtract 107° 25′ 15′′ from 180°; the remainder 72° 34′ 45′′ the angle C., Therefore,

The angle A=45° 41′ 48′′

The angle B=61° 43′ 27′′ Answer.
The angle C=72° 34′ 45′′

Examples for practice, Art. 154.

Example 1. By THEOREM I. CASE I. Angle A+angle B=180°-angle C; that is 54° 30′+63° 10′=180°-angle C, and by transposing and uniting, 180°-117° 40′ 62° 20′ =angle C. To find the side b, sin A: a: :sin B: b; andby the tables, 9.9106860 2.2148438::9.9505223: 2.2546801= 179.754. To find the side c, sin A; a::sin C: c; and, by the tables, 9.9106860 2.2148438: 9.9472689: 2.2514267 =178.4137. Therefore, the angle C=62° 20′, the side b =179.754, and the side c=178.4137.

Example 2. By THEOREM I. CASE II. Figure 27. To find the angle B, a; sin A::b; sin B; and, by the tables. 1.96848299.8502417::2.03342389.9151826-55° 20' 41" the angle B, which added to 45° 6', the angle A, becomes 100° 26′ 41′′, and this subtracted from 180°, leaves 79° 33′ 19′′, the angle C. To find the side c, sin A ;a:: sin Cc; and, by the tables, 9.8502417 1.9684829:: 9.9927439: 2.1109851=129.1-the side c. Therefore the angle B 55° 20′ 41′′, the angle C=79° 33′ 19′′, and the side c=129.1

Example 3. CASE II. To find the angle B, asin A:: b; sin B; and, by the tables. 1.79239179.9653006:: 1.6627578 9.8356667-43° 13′ 58′′. To find the side c, sin A asin Cc; and, by the tables. 9.9653006: 1.79239179.9712100: 1,7983011=62.85=the side c. To find the remaining angle C, subtract the angle A+the angle B 67° 24'+43°13′ 58′′-110° 37′ 58′′, from 180°; the remainder 69° 22′ 2′′ the angle C. Therefore the angle B=43° 13′ 58′′, the angle C 69° 22′ 2′′, and the side c=62.85,

Example 4. CASE II. To find the angle B, asin A::b sin B; and, by the tables. 2.5809250 9.8982992:: 2.2648178 9.5821920=22° 27′ 53′′ the angle B.. The angle A=127° 42′ which added to 22° 27′ 53′′, the angle B, becomes 150° 9′ 53′′, and this subtracted from 180°, leaves 29° 50′ 7′′, the angle C. To find the side c, sin A:a::sin Cc; and by the tables, 9.8982992: 2.5809250 :: 9.69679652.3794223=239.6=the side c. Therefore, the angle B=22°27′ 53′′, the angle C=29° 50′ 7′′, and the side c=239.6.

Example 5. CASE III. As b=58, and c=67, their sum =125, and their difference 9. To find the angles B, C, by THEOREM II, (Art. 144,) (b+c): (b-c)::tan B+C: tan B C; and, by the tables, 2.0969100 0.9542425:: 10.4882240 9.345565-12° 29′ 40′′ the difference of the angles B and C. Their sum =180°-36°=144°, and one half their sum=72°; therefore, the greater angle=84° 29' 40" C, and the less=59° 30′ 20′′=B. To find the side a, sin Bb: sin Aa; and, by the tables, 9.9353452: 1.7634280: 9.76921871.5973015=39.57. Therefore, the angle B=59° 30′ 20′′, the angle C=84° 29′ 40′′, and the side a 39.57.

Example 6. CASE IV. Figure 31. THEOREM III. AB: AC+BC: AC-BC: AP-BP. Substituting numbers, 631814:278; AP-BP; and by taking the logarithms of the numbers, 2.8000294: 2.9106244: 2.4440448 :f 2.5546398=358.62436 the difference of the segments or the longest side; and as the longest side=631, the greate segment, (Alg. 341,)=494.812, and the less=136.18782. To find the angle A, AC R::AP: cos A; and, by the tables, 2.7371926 10.0000000 :: 2.6944297: 9.9572371= 25° 0′ 38′′ the angle A. To find the angle B, BC; R:: BP: cos B; and, by the tables, 2.4281348: 10.0000000 :: 2.1338581 9.7057233 59° 28′ 49′′ the angle B : therefore,

The angle A=25° 0′ 38′′

The angle B=59° 28′ 49′′ Answer.
The angle C-95° 30′ 33′′