(2) One solution when: (a) Angle is acute and opposite side is equal to adjacent side times the sine of the angle. This gives a right triangle. (b) Angle of any size and opposite side greater than adjacent side. (3) Two solutions when angle is acute and the opposite side greater than the adjacent side times the sine of the angle, and less than the adjacent side. The ambiguity of (3) is also apparent from the solution of y found from c sin a the relation sin y = This equation has two values of y less than 180° a each of which may enter into the triangle when a is acute. With each of these values of y there may be found values of 8 and b, thus making two triangles. When logarithms are used, proper conclusions can be drawn from the following, where a, b, and a are given. For other given parts, the proper change can easily be made. If log sin 8 = 0, sin ẞ = 1, B = 90°; hence a right triangle. > 1 which is impossible; hence no solution. If log sin ẞ < 0 and b < a, and therefore ẞ <a, only the acute value of ß can be used; hence there is one solution. If log sin ẞ <0 and b > a, both acute value of ẞ and its supplement may be used; hence there are two solutions. If the given parts are a, c, and a, with a acute and a < c, the formulas for the solution are: The area K can be determined as follows: suppose b, c, and y b sin y = c where ẞ can be determined from sin ß Example. Solve the triangle when a = 11.75, c 15.61, and = a = 34° 15.3'. Solution. Here a is acute, a < c, and a > c sin a, hence there are two solutions. log sin y = log c+ log sin a +colog a = log sin y'. log b' = log a + log sin ẞ'+colog sin a. Let the given 80. Case III. The solution of a triangle when two sides and the included angle are given. First method. parts be a, b, and y. Then from the cosine law, The area K = hb ab sin y; or, in words, the area equals one half the product of the two sides and the sine of the included angle. It is evident that the formula for finding c is not adapted to the use of logarithms. However, this method is often convenient when the numbers expressing the sides contain few figures or when only the third side is to be found. EXERCISES Solve the following triangles by the formulas of this article without logarithms. 81. Case III. Second method. For a solution by logarithms the following theorem is needed: TANGENT THEOREM. In any triangle the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. (a — ẞ) when Formula [37] or [38] makes it possible to find a, b, and y are given, while (a + ẞ) can readily be found because y, therefore a and ẞ can be found from the (a + B) = 90° It is evident now that the other side can be found by the sine theorem, which may also be used as a check. A discussion similar to the above can be given when any two sides and the included angle are given. A convenient set of formulas for solving the triangie when a, b, and y are given is It should be noted that negatives are avoided if the larger angle and side come first in [38]. write [38] in form tan (8 Thus, if ẞ> a and hence b > a, 1 b+ a 2 [38] for finding tan (a - Y); = Example. Solve the triangle when a = 42.367, c 58.964, and |