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Using the Law of Sines and the Law of Cosines to Solve an Oblique Triangle Oblique Triangle c a Law of Sines sin sin sin a b c Law of Cosines a 2 b 2 c 2 2bc cos b 2 a 2 c 2 2ac cos c 2 a 2 b 2 2ab cos b An oblique triangle is a triangle in which none of the angles are right angles. An oblique triangle will either have three acute angles or, two acute angles and one obtuse angle. Case 1: SAA or ASA – One side and two angles are known. One Triangle Given: ( , , and a) OR ( , , and a) OR (, , and b) SAA: ( , , and b) OR ( , , and c) OR (, , and c) ASA: ( , , and c) OR ( , , and b) OR (, , and a) First, subtract from 180° to solve for the unknown third angle. Then, use the Law of Sines TWICE to solve for the remaining two sides. Case 2: SSA 1 – Two sides and the angle opposite one of them are known. Ambiguous Case One Triangle, Two Triangles, or No Triangle Given: (a, b, and ) OR (a, c, and ) OR (b, c, and ) SSA: (a, b, and ) OR (a, c, and ) OR (b, c, and ) First, use the Law of Sines to solve for one of the unknown angles. o If sin = 1, then = 90°, and one right triangle results. (“One Triangle” Case) o If sin > 1, then no triangle results. (“No Triangle” Case) o If sin < 1, then there are TWO values of that satisfy this relationship. The first value of 1 will be in Quadrant I generated directly by the sin-1 key. The second value of 2 will be in Quadrant II and is found by subtracting 1 from 180°. Thus, possibly TWO oblique triangles result. (“One Triangle” or “Two Triangle” Case)2 Second, use the computed value for and the given angle to subtract from 180° to find the unknown third angle. (When sin < 1 there will be two computed values of to consider, and you will need to subtract from 180° twice, once for each value of .) o If you can not find the third angle because the sum of the angles would be > 180°, then no triangle results for that value of . Then use the Law of Sines to solve for the remaining third side. (Use the “Given” AS pair and the “computed” value that you just found for the third angle.) Case 3: SAS – Two sides and the included angle are known. One Triangle Given: (a, b, and ) OR (a, c, and ) OR (b, c, and ) First, use the Law of Cosines to solve for the unknown side. Then, use the Law of Cosines to solve for one of the two unknown angles. 3 Last, subtract from 180° to find the remaining third angle. (Or, use the Law of Cosines again.) Case 4: SSS – Three sides are known. One Triangle Given: (a, b, and c) First, use the Law of Cosines TWICE to solve for TWO of the three unknown angles. Then, subtract from 180° to find the remaining third angle. (Or, use the Law of Cosines again.) Foot Notes: 1) If the acronym spells out a bad word, but backwards, (SSA ASS) then that is the ambiguous case that can result in three possible outcomes: One Triangle, Two Triangles, or No Triangle. One Triangle – This can be generated one of two ways: When sin = 1, then = 90°, and one right triangle results When sin < 1, then there are two values of to consider. If one of the two possible values of results in an impossible triangle then one value of remains, resulting in one oblique triangle. Two Triangles – When sin < 1, then there are two values of to consider. If both values of result in triangles, then two oblique triangle result. No Triangle – When sin > 1, no triangle results since the sine function has range: sin 1 , so that the equation sin = a, has no solution for a > 1. 2) If sin > 0, 0° < < 180°, 90° then could either lie in Quadrant I or Quadrant II. Thus, there are TWO angles to consider. Your calculator only gives you the value for in Quadrant I because of how the range of the Inverse Sine function (from the sin-1 key) is defined. To find the second angle located in Quadrant II you must find the angle in Quadrant II that has the same reference angle as the angle in Quadrant I. To do this subtract the acute angle, , from 180° to obtain the corresponding angle in Quadrant II.. For example, let’s says that sin = 0.8, then = sin–1(0.8), so that = 53.13010235… Thus 1 = 53.1° and 2 = 180° – 53.1° = 126.9°. Now, there are two triangle possibilities to consider one corresponding to each value of . 3) It would have been simpler computationally to use the Law of Sines here, but for sin < 1 there are TWO possible solutions that need to be investigated, while the Law of Cosines results in ONE unique solution. The Law of Cosines produces one unique solution for cos > 0, 0° < < 180°, because only angles in Quadrant I satisfy these two conditions. This corresponds to the same answer generated by your calculator when you use the cos-1 key. Space for additional notes: