... 9.91186 to sin DCB 56° 39′ Then 90-56° 39′-33° 21' B. Hence C DCA+DCB=57° 45'+56° 39′-114° 24'. By the Scale and Compasses. First. Extend the compasses from 324.68 to 193.12 on the line of numbers; that extent will reach from 2.88 to 1.713, the distance of the perpendicular from the middle of the base. Secondly. Extend the compasses from 82.883 on the line of numbers; that extent will reach from 90° to 57° 45′ on the line of sines. Thirdly. Extend the compasses from 95.12 to 79.457 on the line of numbers; that extent will reach from 90° to 56° 39′ on the line of sines. 4. Given the angles and one side of a plane obliqueangled triangle, to find the other two 180°-(A+B)=180-146° 39′-33° 21'= C. By the Scale and Compasses. Extend the compasses from 33° 21' to 65° 36' (the supplement of 114° 24') on the line of sines; that extent will reach from 98 to 162 on the line of numbers. Of the six quantities composing an oblique-angled triangle, it is not necessary to consider only five; because any one of the angles is the supplement of the sum of the other two. Of the three sides and two angles, any three being given, the other two may be found. They may all be reduced to the following cases: 1o. Given the two sides and the angle opposite to one of them. 2o. Given two sides and the angle included by them. 3o. Given the three sides. 4°. Given two angles and the side opposite to one of them. 5o. Given two angles and the side between them. The two last cases may be always reduced to one ; since by having two angles, the third may be found: hence the three angles and a side opposite to one of them are given. From which it appears that the above E five cases may be reduced to four, the data in the two last being "given the three angles and one side." Representing the sides and angles, as before, by a, b, c, and A, B, C; the following equations will be sufficient to solve all the cases of oblique-angled triangles. Sine Casine Axc......(1) a Sine Axb=sine Bya......(3) Sine Axcsine Cxa......(4) Sine Bxa=sine AXb......(5) In the above equations, if the three* angles and any one side be given, the other two sides may be found. Or, if any two sides and an angle opposite to one of them be given, the other angles and the third side may be found. This last includes the ambiguous case already noticed. See Prop. 5, sec. 3. a b tan. (AB)=. cot. C (by Prop. 5, sec. 3.) Then (AB+(90° C) = angle opposite the greater side. And (90°-C)-(AB)= angle opposite the less side. See Prop. 5, sec. 3, note. Or, c=[a2+b2—(24xx cos. C)]. See Prop. 6, sec. 3. Here, two sides and the contained angle are given, from which the third side and the two remaining angles are obtained. * When two angles of a triangle are given, the third is given, as was before observed. In these equations, the sides are given, from which the respective angles are found. See Prop. 6, sec. 3. The above formulas are not adapted to logarithmic solutions, being composed of additive and subtractive members. Exercises in Right-angled Triangles. 1. In a right-angled triangle, given the base AB 180 yards, and the angle A 62° 40'; required the hypothenuse AC, and the perpendicular CB. 2. In the triangle ABC, given AB 160 perches, and the angle A 56° 30'; required the perpendicular BC. Answ. 241.7 perches. 3. In the triangle ABC, given AC 5 feet, and AB 3 feet; required the side BC. Answ. 4 feet. 4. In the triangle ABC, given AB 300, BC 400; required the side AC. Answ. 500. 5. In the triangle ABC, there is given AB 100 miles, and the angle A 47° 30′; required the side BC. Answ. 109.13 miles. 6. In the triangle ABC, there is given AB 18 yards, and the angle A 62° 40'; required the sides BC and AC. Answ. BC 34.82 yards, and AC 39.2 yards. 7. In the triangle ABC, there is given AC 39.2 yards, and AB 18; required the side BC and the angle A. Answ. BC 34.82, and the angle A 62° 40′. Exercises in Oblique-angled Triangles. 1. In the triangle ABC, there is given AB 360, the angle A 43° 15', the angle B 72° 51′, and the angle C 63° 54′; required the sides AC, and CB. Answ. AC 383.1, and BC 274.7. 2. In the triangle ABC, there is given the side AB 365 yards, the angle A 57° 12′, and the angle B 24° 45'; required the angle C, and the sides AC and BC. Answ. The angle C 98° 3', the side AC 154.33 yards, and BC 309.86 yards. 3. In the triangle ABC, there is given AC 120 poles, |