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The Law of Sines We know that Trigonometry can help us solve right triangles. But not all triangles are right triangles. Fortunately, Trigonometry can help us solve non-right triangles as well. Non-right triangles are know as oblique triangles. There are two categories of oblique triangles—acute and obtuse. Acute Triangles In an acute triangle, each of the angles is less than 90º. Obtuse Triangles In an obtuse triangle, one of the angles is obtuse (between 90º and 180º). Can there be two obtuse angles in a triangle? Of course not! Law of Sines sin( ) sin( ) sin( ) a b c The Law of Sines is used when we know any two angles and one side or when we know two sides and an angle opposite one of those sides. There are two cases where one side and two angles are known: ASA or SAA Disclaimer: Do not assume that any triangles are drawn to scale. Let’s do an example involving ASA. From the model, we need to determine a, b, and using the law of sines. First off, 42º + 61º + = 180º so that = 77º. (Knowledge of two angles yields the third!) Now by the law of sines, we have the following relationships: sin(42 ) sin(77 ) sin(61 ) sin(77 ) ; a 12 b 12 Let’s solve for our unknowns: o 12sin(42 ) a o sin(77 ) 12(0.6691) a 0.9744 a 8.2401 o 12sin(61 ) b o sin(77 ) 12(0.8746) b 0.9744 b 10.7709 Now, an example involving SAA: From the model, we need to determine a, b, and using the law of sines. Note: + 110º + 40º = 180º so that = 30º b a By the law of sines, we have the following relationships: sin(30 ) sin(40 ) sin(110 ) sin(40 ) ; a 12 b 12 Therefore, o 12sin(30 ) a o sin(40 ) 12(0.5) a 0.6428 a 9.3341 o 12sin(110 ) b o sin(40 ) 12(0.9397) b 0.6427 b 17.5428 The Ambiguous Case – SSA In this case, you may have information that results in one triangle, two triangles, or no triangles. Example #1 of SSA Two sides and an angle opposite one of the sides are given. Let’s try to solve this triangle. By the law of sines, sin(57 ) sin( ) 15 20 Thus, o 20sin(57 ) sin( ) 15 20(0.8387) sin( ) 15 sin( ) 1.1183 Impossible! Therefore, there is no value for that exists! No triangle is possible! Example #2 of SSA Two sides and an angle opposite one of the sides are given. Let’s try to solve this triangle. By the law of sines, sin(32 ) sin( ) 30 42 So that, o 42sin(32 ) sin( ) 30 42(0.5299) sin( ) 30 sin( ) 0.7419 48 or 132 o o Interesting! Let’s see if one or both of these angle measures makes sense. Case 1 Case 2 132 32 180 48 32 180 16 100 Both triangles are valid! Therefore, we have two possible cases to solve. Finish Case 1: sin(100 ) sin(32 ) c 30 o 30sin(100 ) c o sin(32 ) 30(0.9848) c 0.5299 c 55.7539 Finish Case 2: sin(16 ) sin(32 ) c 30 o 30sin(16 ) c o sin(32 ) 30(0.2756) c 0.5299 c 15.6029 Wrapping it up, here are our two solutions: Example #3 of SSA: Two sides and an angle opposite one of the sides are given. Let’s try to solve this triangle. By the law of sines, sin(40 ) sin( ) 3 2 o 2sin(40 ) sin( ) 3 2(0.6428) sin( ) 3 sin( ) 0.4285 25.4 or 154.6 o o Note: Only one is legitimate! 40 25.4 180 114.6 40 154.6 180 14.6 Not Possible! Thus, we have only one triangle. Now let’s find b. By the law of sines, o o sin(114.6 ) sin(40 ) b 3 o 3sin(114.6 ) b o sin(40 ) 3(0.9092) b 0.6428 b 4.2433 Finally, we have: A road slopes 15 above the horizontal, and a vertical telephone pole stands beside the road. The angle of elevation of the Sun is 65 , and the pole casts a 15 foot shadow downhill along the road. Find the height of the pole. Let x the height of the pole. A 65º BAC 180 90 65 25 ACB 65 15 50 x B sin 25 sin 50 15º 15 x C 15sin 50 x 27.2 sin 25 The height of the pole is about 27.2 feet. o o o o o o o We can find the area of a triangle if we are given any two sides of a triangle and the measure of the included angle. (SAS) 1 Area = (side)(side)(sine of included angle) 2 Example: Find the area of ABC given a = 32 m, b = 9 m, and mC 36. 1 Area 32m 9m sin36 2 Area 84.6m 2