Extend from 46° 30', to 72° 59', on the line of sines; that distance will reach from 200 to 263,7 on the line of numbers, for AC. Note. Tlie method by Natural Sines will be obvious from the foregoing analogies. CASE II. Two angles and a side given; to find the other sides PL. 5. fig. 15. In the triangle ABC, there is the angle A 46° 30' AB 230; and the angle B 37° 30', given to find AC and BC. 1st. By Construction. Draw a blank line, upon which set AB 230, from a scale of equal parts ; at the point A of the line AB, make an angle of 46° 30, by a blank line ; and at the point B of the line AB make an angle of 37° 30', by another blank line : the intersection of those lines gives the point C, then the triangle ABC is constructed. Measure AC and BC from the same scale of equal parts that AB was taken ; and you have the answer required. 2d. By Calculation. By (cor. 1. theo, 5. sect. 4.) 180°— the sum of the angles A and B=C. A 46° 30' 180° ---- 84.00/=96° 00'=C. By def. 27. sect. 4. The sine of 96°=the sine of 84', which is the supplement thereof; therefore instead of the sine of 96', look in the tables for the sine of 84o. Extend from 84' (which is the supplement of 969) to 46° 30' on the sines; that distance will reach from 230 to 168, on the line of numbers, for BC. Extend from 84° to 37o. 30', on the sines; that extent will reach from 230 to 141, on the line of numbers, for AC. CASE III. I'wo sides and a contained angle given; to find the other anglea and side. Pl. 5. fig. 16. In the triangle ABC, there is AB 240, the angle A 36° 40' and 40 180, gtven ; to find the angles and B, and the side BG. 1st. By Construction. Draw a blank line, on which from a scale of equal parts, lay AB 240; at the point A of the line AB, make an angle of 36° 40', by a blank line ; on which from A, lay AC 180, from the same scale of equal parts; measure the angles C and B, and the side BC, as before; and you have the answers required. 2d. By Calculation. By cor. 1. theo. 5. sect. 4. 180° — the angle A 36. 40' = 143°. 20' the sum of the angles C and. B: therefore half of 143. 20', will be half the sum of the two required angles, C and B. By theo. 2. of this sect. As the sum of the two sides AB and AC=420 is to their difference, = 60 71° 40 So is the tangent of half the sum of the two unknown angles Cand B 23° 20 1 By theo. 4. To half the sum of the angles C and B=71° 40% Add half their difference as now found = 23 20. The sum is the greatest angle, or ang. C=95 00 Subtract, and you have the least angle, or B=4820 The angle C and B being found; BC is had, as before, by theo. 1. of this sect. Thus, 8. B : AC::8: A: BC. 3d. By Gunter's Scale. Because the two first terms are of the same kind, extend from 420 to 60 on the line of numbers ; lay that extent from 45° on the line of tangents, and keeping the left leg of your compasses fixed, move the right leg to 71°. 40';, that distance laid from 45° on the same line will reach to 23o. 30', the half difference of the required angles. Whence the angles are obtained, as before. The second proportion may be easily extended, from what has been already said. CASE IV. PL. 5. fig. 17. The three sides given, to find the angles. In the triangle ABC, there is given, AB 64, AG 47, BC 34; the angles A, B, C, are required. Ist. By Construction. The construction of this triangle must be manifest, from prob. 1. sect. 4. 2d. By Calculation. From the point C, let fall the perpendicular CD on the base AB; and it will divide the triangle into two right angled ones, ADC and CBD; as well as the base ÅB, into the two segments, AD and DB AC 47 Sum 81 Difference 13 By theo. 3. of this sect. As the base or the longest side, AB 64 is to the sum of the other sides, AC and BC, 81 So is the difference of those sides 13 to the difference of the segments of 16.46 the base AD DB. of} By theo. 4. of this sect. 32 To half the base, or to half the sum of the segments AD and DB. Add half their difference, now found, 8.23 |