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The Mathematics 11 Competency Test Solving Right Triangles As mentioned already, to solve a right triangle means to calculate all unknown lengths and angles. Now, every triangle (including right triangles) have three angles and three sides, for a total of six parts (as they are called). To solve a triangle, you need to know three of the six parts, and the three known parts must include the length of at least one side. Since we always know at least one angle of a right triangle (the right angle itself), then the minimum additional information needed to solve a right triangle is either (i) the length of two of the three sides (called the ‘ss’ case) or (ii) the length of one side and the value of one of the acute angles (called the ‘sa’ case) Some books give formal strategies for each type of right triangle problem. Here, we just advise the following procedure: (i) list the values of the parts that are known (you don’t need to list the right angle) (ii) list the symbols for the parts which are unknown and must be calculated. (iii) then, use Pythagoras’s Theorem and/or the definitions of the principal trigonometric functions and the inverse trigonometric functions to calculate each of the unknown parts, one-by-one. It is a good strategy to set up your calculations so that they use just the original known values given in the statement of the problem. Then if you make an arithmetic error, it doesn’t affect other calculations. B Example 1: Solve the right triangle shown in the sketch. solution: We are given 31 42 a = 31 and c = 42 and so we must determine b, A, and B. When two sides of a right triangle are given, the third side can always be determined using Pythagoras’s Theorem: A b2 = c 2 – a 2 = 422 – 312 = 803. Thus b= 803 ≅ 28.34 rounded to two decimal places. Then David W. Sabo (2003) Solving Right Triangles Page 1 of 4 sin A = 31 , 42 so 31 0 A = sin−1 ≅ 47.57 42 Finally, cos B = 31 42 and so 31 B = cos−1 ≅ 42.430 42 Thus the required solution is b ≅ 28.34, A ≅ 47.570, and B ≅ 42.430. Notice how we organized our work in this example so that all calculations were based only on information given in the original problem. None of the calculations involved numbers we had previously calculated ourselves. This is a good strategy, because then if we had made an arithmetic error in one step of the solution, it wouldn’t affect the correctness of results of other calculations. This sort of defensive strategy is always possible when solving right triangles. Example 2: Solve the right triangle sketched to the right. B solution: We are given A = 270 and b = 412 m To solve this triangle, we need to compute values for B, a, and c. 270 When one of the acute angles is given, the easiest way to compute the other is by subtraction from 900: A + B = 900 So A (for a right triangle) 270 + B = 900 giving B = 900 – 270 = 630. David W. Sabo (2003) Solving Right Triangles Page 2 of 4 Then, for side a, we have tan A = a b or tan270 = a 412 m which gives ( ) a = ( 412 m ) tan270 ≅ 209.92 m . Finally, for side c, cos A = b c or cos 270 = 412 m c giving c= 412 m ≅ 462.40 m cos 270 Thus, the required solution is: B = 630, a ≅ 209.92 m, c ≅ 462.40 m. Again, note the strategy employed in the previous example. To calculate ‘a’, we looked for a trigonometric ratio that involved A and b (our given quantities) and ‘a’ (the unknown we wished to determine). Similarly, to determine c, we looked for a trigonometric ratio involving A, b, and c. In both of these cases this meant that the definition of the selected trigonometric ratio would amount to an equation with one unknown, which was then easily solved. Example 3: Solve the right triangle shown in the figure to the right. A 520 solution: The solution here is that there is no solution. Knowing all three angles of a right triangle is not enough information to be able to calculate the lengths of any of the three sides. In fact, there is an infinite number of right triangles that have these three angles – each of a different size. David W. Sabo (2003) 380 B Solving Right Triangles Page 3 of 4 So, we cannot obtain a unique solution for this triangle. Example 4: Solve the right triangle with c = 26.8 cm and B = 37.560. solution: It is probably best to start by making a sketch, as is done to the right. A In addition to the right angle, of course, the known parts of the triangle, B and c, are given in the statement of the problem. So, we need to determine the values of A, a, and b. b First, A + B = 90 so 37.560 0 0 B A + 37.56 = 90 a C 0 giving A = 900 – 37.560 = 52.440 Then sin B = b c gives sin37.560 = or b 26.8 cm b = (26.8 cm)(sin 37.560) ≅ 16.34 cm Finally, using cos B = a c or cos37.560 = a 26.8 cm gives us a = (26.8 cm)(cos 37.560) ≅ 21.24 cm Thus, the required solution is; A = 52.440, a ≅ 21.24 cm, b ≅ 16.34 cm. David W. Sabo (2003) Solving Right Triangles Page 4 of 4