AD: BE :: DF: BF; that is, A+B A-B 2 2 THEOREM III. (51.) If from any angle of a triangle a perpendicular de drawn to the opposite side or base, the whole base will be to the sum of the other two sides, as the difference of those two sides is to the difference of the segments of the base. For demonstration, see Geometry, Prop. 31, Cor., B. IV. (52). In every plane triangle, three parts must be given to enable us to determine the others; and of the given parts, one, at least, must be a side. For if the angles only are given, these might belong to an infinite number of different triangles In solving oblique-angled triangles, four different cases may therefore be presented. There may be given, 1. Two angles and a side; We shall represent the three angles of the proposed triangle by A, B, C, and the sides opposite them, respectively, by a, b, c. CASE I. The pro (53.) Given two angles and a side, to find the third angle and the other two sides. To find the third angle, add the given angles together, and subtract their sum from 180°. The required sides may be found by Theorem I. The portion will be The sine of the angle opposite the given side: the given side :: the sine of the angle opposite the required side : the re quired side. Ex. 1. In the triangle ABC, there are C given the angle A, 57° 15', the angle B, 35° 30', and the side c, 364, to find the other parts. The sum of the given angles, subtracted A a from 180°, leaves 87° 15' for the angle C. Then, to find the side a, we say, sin. C:c:: sin. A:a. By natural numbers, .998848 : 364 ::.841039 : 306.49=a. This proportion is most easily worked by logarithms, thus : As the sine of the angle C, 87° 15', comp., 0.000500 2.561101 2.486417. To find the side b: sin. C;c:: sin. B : b. By natural numbers, .998848 : 364 ::.580703 : 211.62=b. The work by logarithms is as follows: sin. C, 87° 15', comp., 0.000500 : C, 364, 2.561101 :: sin. B, 35° 30', 9.763954 :b, 211.62, 2.325555. Ex. 2. In the triangle ABC, there are given the angle A, 49° 25', the angle C, 63° 48', and the side c, 275, to find the other parts. Ans., B=66° 47'; a=232.766; b=281.67. CASE II. (54.) Given two sides and an angle opposite one of them, to find the third side and the remaining angles. One of the required angles is found by Theorem I. The proportion is, The side opposite the given angle : the sine of that angle :: the other given side : the sine of the opposite angle. The third angle is found by subtracting the sum of the other two from 180° ; and the third side is found as in Case I. If the side BC, opposite the given an С gle A, is shorter than the other given side AC, the solution will be ambiguous; that is, two different triangles, ABC, AB'C, may be formed, each of which will satisfy Á B the conditions of the problem. The numerical result is also ambiguous, for the fourth term B B' of the first proportion is a sine of an angle. But this may be the sine either of the acute angle AB'C, or с of its supplement, the obtuse angle ABC (Art. 27). In practice, however, there will generally be some circumstance to determine whether the required angle is acute or A B obtuse. If the given angle is obtuse, there can be no ambiguity in the solution, for then the remaining angles must of course be acute. Ex. 1. In a triangle, ABC, there are given AC, 458, BC, 307, and the angle A, 28° 45', to find the other parts. To find the angle B: BC : sin. A::AC: sin. B. By natural numbers, 307 : .480989 :: 458 : .717566, sin. B, the arc corresponding to which is 45° 51' 14", or 134° 8' 46". This proportion is most easily worked by logarithms, thus · BC, 307, comp., 7.512862 : sin. A, 28° 45', 9.682135 :: AC, 458, 2.660865 : sin. B, 45° 51' 14", or 134° 8' 46'', 9.855862. The angle ABC is 134° 8' 46', and the angle AB'C, 45° 51' 14". Hence the angle ACB is 17° 6' 14", and the angle ACB', 105° 23' 46". To find the side AB: sin. A: CB :: sin. ACB : AB. By logarithms, sin. A, 28° 45', comp., 0.317865 : CB, 307, 2.487138 :: sin. ACB, 17° 6' 14", 9.468502 : AB, 187.72, 2.273505. To find the side AB': sin. A: CB' :: sin. ACB' : AB'. By logarithms, sin. A, 28° 45', comp., 0.317865 : CB', 307, 2.487138 :: sin. ACB', 105° 23' 46', 9.984128 : AB', 615.36, 2.789131. Ex. 2. In a triangle, ABC, there are given AB, 532, BC, 358, and the angle C, 107° 40', to find the other parts. Ans. A=39° 52' 52" ; B=32° 27' 8" ; AC=299.6. In this example there is no ambiguity, because the giver. angle is obtuse. Case III. (55.) Given two sides and the included angle, to find the third side and the remaining angles. The sum of the required angles is found by subtracting the given angle from 180°. The difference of the required angles is then found by Theorem II. Half the difference added to half the sum gives the greater angle, and, subtracted, gives the less angle. The third side is then found by Theorem I. Ex. 1. In the triangle ABC, the angle A is given 53° 8'; the side c, 420, and the side b, 535, to find the remaining parts. The sum of the angles B+C=180°-53° 8'=126° 52'. Half their sum is 63° 26'. Then, by Theorem II., 535+420 : 535-420 :: tang. 63° 26' : tang. 13° 32 25", which is half the difference of the two required angles. Hence the angle B is 76° 58' 25', and the angle C, 49° 53' 35". To find the side a: sin. C:C:: sin. A:a=439.32. Ex. 2. Given the side c, 176, a, 133, and the included angle B, 73°, to find the remaining parts. Ans., b=187.022, the angle C, 64° 9' 3'', and A, 42° 50' 57". CASE IV. (56.) Given the three sides, to find the angles. Let fall a perpendicular upon the longest side from the opposite angle, dividing the given triangle into two right-angled triangles. The two segments of the base may be found by Theorem III. There will then be given the hypothenuse and one side of a right-angled triangle to find the anglis. Ex. 1. In the triangle ABC, the side a is 261, the side by 345, and c, 395. What are the angles ? Let fall the perpendicular CD upon AB. . or a Then, by Theorem III., AB : AC+CB :: AC-CB: AD-DB; 395 : 606 :: 84 : 128.87. Half the difference of the segments added to half their sum gives the greater segment, and subtracted gives the less segment. Therefore, AD is 261.935, and BD, 133.065. Then, in each of the right-angled triangles, ACD, BCD, we have given the A D B В hypothenuse and base, to find the angles by Case II. of rightangled triangles. Hence AC:R:: AD: cos. A=40° 36' 13"; BC:R:: BD : cos. B=59° 20' 52". Therefore the angle C=80° 2' 55". Ex. 2. If the three sides of a triangle are 150, 140, and 130, what are the angles ? Ans., 67° 22' 48', 59° 29' 23', and 53° 7' 49" Examples for Practice. 1. Given two sides of a triangle, 478 and 567, and the in cluded angle, 47° 30', to find the remaining parts. 2. Given the angle A, 56° 34', the opposite side, a, 735, and the side b, 576, to find the remaining parts. 3. Given the angle A, 65° 40', the angle B, 74° 20', and the side a, 275, to find the remaining parts. 4. Given the three sides, 742, 657, and 379, to find the angles. 5. Given the angle A, 116° 32', the opposite side, a, 492, and the side c, 295, to find the remaining parts. 6. Given the angle C, 56° 18', the opposite side, c, 184, and the side b, 219, to find the remaining parts. This problem admits of two answers. INSTRUMENTS USED IN DRAWING. (57.) The following are some of the most important instruments used in drawing. 1. The dividers consist of two legs, revolving upon a pivot at one extremity. The joints should be composed of two dif |