263. The same results may be obtained geometri cally. Construction. Draw AB= c; make the angle ABD = the given angle B; with centre A and radius = b describe a circle; draw AD perpendicular to BD. 1. Ifb is less than c sin B, i. e. less than AD, the circle will not cut BD at all, and the construction fails. (Fig. i.) 2. If 6 is equal to AD, the circle will touch the line BD in the point D, and the required triangle is the right-angled triangle ABD. (Fig. ii.) 3. If b is greater than AD and less than AB, i. e. than c, the circle will cut the line BD in two points C1, C, each on the same side of B. And we get two triangles ABC,, ABC, each satisfying the given condition. (Fig. iii.) 4. If b is equal to c, the circle cuts BD in B and in one other point C; if b is greater than c the circle cuts BD in two points, but on opposite sides of B. In either case there is only one triangle satisfying the given condition. (Fig. iv.) In this b, c, B are given, a is unknown. Write x for a and we get the quadratic equation x2- 2c cos B. x = b2 — c2. Whence, x2 - 2c cos B. x + c2 cos2 B = b2 - c2 + c2 cos3 B = b2 – c2 sin3 B ; .. x = ccos B± √b2 – c2 sin3 B. Let a,, a, be the two values of x thus obtained, then a1 = c cos B+ √b2 – c2 sin2 B Which of these two solutions is admissible, may be decided as follows; 1. Ifb is less than c sin B, then (b2 - c2 sin3 B) is negative, so that a, a, are impossible quantities. 2. If b is equal to c sin B, then (b3 — c3 sin3 B) = 0, and a1 = a,; thus the two solutions become one. 3. If b is greater than c sin B, then the two values a a are different and positive unless 4. If b is equal to c, then a, = 0; if b is greater than c, then a, is negative and is therefore inadmissible. In either of these cases a, is the only available solution. 265. We give two examples. In the first there are two solutions, in the second there is only one. Example 1. Find A and C, having given that b=379-41 chains, c-483-74 chains, and B=34° 11'. L sin C-log c+L sin B-log b =2.6846120+9.7496148-2.5791088 =9.8551180=L sin 45° 45'; .. C=45° 45', or, 180°-45° 45′ =134° 15′. Since b is less than c, each of these values is admissible. When C 45° 45', then A-100° 4′. When C-134° 15′, then A=11° 34'. Example 2. Find A and C, when b=483 74 chains, c=379.14 chains, and B=34° 11′. L sin C-log c+L sin B-log b =2.5791088+9.7496148-2-6846120 .. C=26° 9′, or, 180o – 26° 9′ = 153° 51′. Since b is greater than c, C must be less than 90°, and the larger value for C is inadmissible. [It is also clear that (153° 51′+34°11′) is >180°]. .. C=26° 9′, A=119° 40'. EXAMPLES. LXIX. (1) If B=40°, b=140·5 feet, a= 170.6 feet, find A and C. log 1.405=1476763, I sin 40° =9.8080675, log 1.706=2319790, L sin 51° 18′ =9.8923342, L sin 51° 19'=9.8924354. (2) Find B and C, having given that A=50o, b=119 chains, a=97 chains, and that (3) Find B, C, and c, having given that A=50°, b=97, a=119 (see example (2)). log 1.553=191169, L sin 38° 38′ 24′′-9.795479, (4) Find A, having given that a=! =24, c=25, C=65° 59′, and that log 2.5=3979400, L sin 65° 59′=9.9606739, L sin 61° 17'=9.9430028. (5) If a=25, c=24, and C=65° 59′, find A, B and the greater value of b. log 1.755=2442771, Z sin 72° 4′ = 9.9783702, L sin 41° 56' 26"-9.8249959; for other logs see example (4). (6) Supposing the data for the solution of a triangle to be as in the three following cases (a), (B), (y), point out whether the solution will be ambiguous or not, and find the third side in the obtuse-angled triangle in the ambiguous case: (a) A=30o, a=125 feet, c=250 feet, (B) A=30o, a=200 feet, c=250 feet, log 2=3010300, L sin 38° 41'=9.7958800, 266. It saves a little trouble in practice when using the formula a= b sin A to write it thus ; a=b sin A. cosec B. For then, log a = log b + L sin A+ L cosec B – 20. Thus the subtraction of a logarithm is avoided. * 267. In the following Examples the student must find the necessary logarithms etc. from the Tables. * MISCELLANEOUS EXAMPLES. LXX. (1) Find A when a=374.5, b=5762, c=759.3 feet. (2) Find B when a=4001, b=9760, c=7942 yards. (3) Find C when a=8761·2, b=7643, c=4693·8 chains. (4) Find B when A=86° 19′, b=4930, c=5471 chains. (5) Find C when B=32° 58′, c=1873·5, a=764·2 chains. (6) Find c when C=108° 27′, a=36541, b=89170 feet. (7) Find c when B=74° 10′, C=62° 45', b=3720 yards. (8) Find 6 when B=100° 19′, C=44° 59′, a=1000 chains. (9) Find a when B=123° 7′ 20′′, C=15° 9′, c=9964 yards. Find the other two angles in the six following triangles. (10) C=100° 37′, b=1450, C= =6374 chains. (11) C=52° 10′, b=643, c=872 chains. (12) A=76° 2′ 30′′, b=1000, a=2000 chains. (13) C=54° 23′, b=8734, c=752.8 feet. (14) C=18° 21′, b=674·5, c=269.7 chains. (15) A=29° 11′ 43′′, b=7934, a=4379 feet. (16) The difference between the angles at the base of a triangle is 17° 48′, and the sides subtending those angles are 105.25 feet and 76.75 feet; find the third angle. (17) Ifb : c=4:5, a= = 1000 yards and A=37° 19′, find b. The student will find some Examples of Solution of Triangles without the aid of logarithms, in an Appendix. |