Given two sides and their included angle, required the remaining parts. Let ABC be a triangle, and let the sides opposite the angles A, B, C, be denoted, respectively, by a, b, C. Let a and b, and their E D HF B included angle C, be given, and the remaining parts, A, B, and c, required. The sum of the angles A and B is found from the formula, A B 180° C. With C as a center, and b, the shorter of the two given sides, as a radius, describe a circumference cutting a in D, a produced in E, and c in H. Draw AE, AD, CH, and DF parallel to AE. The angle DAE is a right angle, since it is inscribed in a semi-circle; hence, its alternate angle, ADF, is also a right angle. The angle ACE being exterior to the triangle ABC, is equal to AB. But ACE having its vertex at the center, is measured by the intercepted arc AE. The inscribed angle ADE is measured by one-half the arc AE; hence, ADE ACE = ¦ (A + B). But HCB, having its vertex at the center, is measured by the intercepted arc DH; and DAF, being an inscribed angle, is measured by one-half the arc DH; hence, DAF HCB = } (A — B). = DF AD tan DAF AD tan (A-B). = = From the similar triangles, ABE and FBD, we have BE: BD :: AE: DF. Since CE CA, BE BC+ CA=a+b. = Since CDCA, BD= BC- CA a - b. Substituting the values of BE, BD, AE, and DF in the above proportion, and omitting the common factor AD in the second couplet, we have a+ba-b: tan (A+B): tan (A—B). Hence, In any plane triangle, the sum of the sides including an angle is to their difference as the tangent of half the sum of the other two angles is to the tangent of half their difference. We find from the proportion, the equation tan (A-B) (ab) tan (A+B) a + b ... log tan 1⁄2 (A——B) = log (a−b) + log tan †(A+B) a. c. log (a+b)-- 10. We have now found (A+B) and 1⁄2 (A — B). A = {(A+B) +}(A−B), B = {(A+B) − 1 (A —- B). sin A sin C::a: c, a sin C sin A .. log clog a + log sin C+ a. c. log sin A-10. S. N. 6. a+bab: tan (A+B): tan (A-B). ... tan (A — B) · (a - b) tan (A+B) a + b log tan (A-B) = log (a—b) + log tan (A+B) A = ¦ (A + B) + 1⁄2 (A — B) = 74° 7′ 25′′. B = į — : sin A sin C: a c, = 3 a sin C log clog a log sin C+ a. c. log sin A-10. Given the three sides of a triangle, required the angles. Let ABC be a triangle, take the longest side for the base, and draw the perpendicular p from the vertex B to the base. A B P a Denote the segments of the base by s and s' respectively. Then, (1) c2-8'2= p2, and (2) a2-82= p2. .'. (3) c2—s′2—a2—s2, .'. (4) s2—s′2—a2—c2. .'. (5) (8+8′) (8 — s′) = (a + c) (a−c). ... (6) s + s'′ : a + c :: a -c: 8-s'. Hence, The sum of the segments of the base is to the sum of the other sides as the difference of those sides is to the difference of the segments. .'. (8) log (s— s′) = log (a+c) -+ log (a−c) a. c. log (88) 10. In case the sides of the triangle are small, find s-s' from (7); otherwise, it will be more convenient to employ (8). |