for solving Case III, the following is selected because it offers some advantages over the second method. Take as the given parts a, b, and y. Choose the auxiliary angle x so that In choosing tan x, put the larger of the given sides in the numerator, and thus make x > 45°. This avoids the negative angle in tan (x - 45°). Example. Solve by this method the example given under the second method. This method can be used to good advantage when the triangle is in a series of triangles where the logarithms of c and a have already been found. This saves finding c + a and c logarithms of them. a and the EXERCISES 1. Given a = 49.366, b = 26.437, and y = 47° 18.6'; find find 2. Given a = find = 55° 41' 26". 137° 8' 49"; c = 36.962, a = 100° 58.3', ẞ = 31° 43.1', and K 3. Given b = 247.81, c = 4. Given a 36.518, b find find C = 42.954, α = 38° 57′ 52", and B = 8° 44' 8". 5. Given a = 67.375, c = 26.858, and B 43.213, α = 147° 11.3', and y = 12° 28.5′. 6. Find the areas of the triangles in Exercises 3 and 5. = 479.65. 7. As in Case III, third method, is it always possible to choose an angle x such that tan x = 응? ? Show why. 83. Case IV. The solution of a triangle when the three sides are given. In this case the angles can be found by means of the cosine theorem, from which the following formulas are derived: These formulas give the cosines of the angles and therefore the angles; but they are not adapted to logarithms and so are convenient only when the sides are expressed in numbers of few figures, or when a table of squares is at hand. 84. Case IV. Formulas adapted to the use of logarithms. (a) Start with the equation cos α = member of it from 1. This gives b2 + c2 - a2 2 bc b2 + c2-a2 and subtract each a2 (b22 bc + c2) 2 bc Let a+b+c=2s. Then ab + c = 2 (sb), and a +b−c = 2 (sc). Substituting these values in the above, (b) By adding each member of the equation cos a = b2 + c2-a2 2 bc to 1, and carrying out the work in a manner similar to the above, there are obtained the following: (c) By dividing each formula of the set under (a) by the corresponding formula of the set under (b) there results These last three can be put in a form slightly more convenient. Since by tan B = tan Y = 8 с In using any of these sets of formulas, the work may be checked * Formula [43] was discovered by Hero (or Heron) of Alexandria about the beginning of the Christian era. Since the sine varies most rapidly for small angles, and the cosine most rapidly for angles near 90°, formulas [39] should be used when the angles are small, and [40] when the angles are near 90°. In all cases the tangent varies more rapidly than either sine or cosine. Hence formulas [41] or [42] are always more nearly accurate than [39] or [40]. Again formulas [41] or [42] are more convenient since, for a complete solution of the triangle, they require only four logarithms to be taken from the table; while [39] and [40] require, respectively, six and seven. Formulas [42] may be derived by taking from geometry the fact that the area of a triangle, when the three sides are given, is |