Survey

# Download Law of Sines and Cosines The entire trigonometry that has been

Document related concepts

Transcript

Law of Sines and Cosines The entire trigonometry that has been done thus far has dealt with the indirect measurement of sides and angles of right triangles. The calculations were carried out using Pythagorean Theorem and/or Trigonometric Ratios. hyp 2 s1 2 s 2 2 sin opp hyp cos adj hyp tan opp adj This process can now be expanded to include oblique triangles. An oblique triangle is one that has no right angle. The three angles can be acute (all angles less than 900) or one angle can be obtuse (greater than 900) and the remaining two angles acute. Regardless of what scenario exists, the measure of unknown angles and the lengths of unknown sides can be determined by using either the Law of Sines or the Law of Cosines. Let us look first at the Law of Cosines. The Law of Cosines can be used to find the length of an unknown side if we know the length of two sides of the triangle and the measure of the angle they form. This angle is called the included angle. Knowing two sides and the included angle is referred to as SAS. Another way to refer to this is a “V”. The lines of the letter “V” represent the sides and the vertex of the “V” is the included angle. Students can use their fingers to determine if these measurements are known. Example 1: Given ABC with b = 9cm, c = 12cm, and A = 620, calculate the length of side ‘a’. Solution 1: When the triangle has been sketched, it is obvious that SAS does exist. To determine the length of side ‘a’ the following formula is used: a b 2 c 2 2bcCosA a 92 122 2912(Cos62 0 ) a 123.5941424 a 11.1cm. Notice that the lower case letters are used to name the sides of the triangle. These lower case letters, used for the sides, are those of the corresponding angles. Once the values have been substituted into the formula, the calculations can be done on the calculator by typing in exactly what is written under the radical sign. Example 2: Given ABC with a = 15m, c = 28m and B = 1120, calculate the length of side ‘b’. Solution 2: When the triangle has been sketched, it is obvious that SAS does exist. To determine the length of side ‘b’ the following formula is used: b a 2 c 2 2acCosB b 152 282 21528Cos112 0 b 1323.669538 b 36.4cm Notice that the letter of the side that is being calculated only appears once in the formula and that is for its corresponding angle. The remainder of the formula consists of the letters of the other two sides – the sum of their squares minus twice their product times the cosine of the included angle. If this highlighted statement is repeated when the formula is used in class, the students will adapt it when using the formula themselves. The Law of Cosines is also used to determine the measure of an unknown angle when the lengths of all the sides of the triangle are known. This is referred to as SSS. Students should be reminded of the fact that the longest side of a right triangle, the hypotenuse, is located opposite the right angle which is the largest angle of the triangle. The same will follow for an oblique triangle – longest side opposite largest angle and shortest side opposite smallest angle. This formula for determining the measure of an angle is the result of solving a 2 b 2 c 2 2bcCosA in terms of CosA. Example 3: Given ABC with a = 18 cm, b = 21 cm, and c = 27 cm, calculate the measure of the smallest angle (nearest tenth) Solution 3: The shortest side is side ‘a’ which makes A the smallest angle. This is the angle we must find. b2 c2 a2 CosA 2bc 2 2 (21) 27 2 18 CosA 22127 846 CosA 1134 CosA .7640 cos 1 CosA cos 1 .7640 A 41.8 0 Notice that the answer was determined by working with the formula as a fraction – result of the numerator over result of the denominator. The other option is to use brackets (numerator) / (denominator). However, students must be very careful when using the brackets since failure to open and close all brackets will result in an error. To err on the side of caution, the first option will not result in an error. Example 4: Given ABC with a = 118 cm, b = 55 cm, and c = 65 cm, calculate the measure of A (nearest tenth). Solution 4: The triangle was sketched as being an obtuse triangle because side ‘a’ was much longer than the other two sides. b2 c2 a2 CosA 2bc 2 2 2 55 65 118 CosA 25565 6674 CosA 7150 CosA .9334 cos 1 CosA cos 1 .9334 A 159 0 CosA is a negative value. This indicates that the angle is greater than 900 since the cosine function is negative in the second quadrant. 90 0 A 180 0 Now that we have seen the Law of Cosines in action, it is time to explore the Law of Sines. In order to use the Law of Sines to determine the length of a side, we must know the measure of an angle and its corresponding side plus the measure of the corresponding angle of the unknown side. To use the Law of Sines to determine the measure of an angle, we must know the measure of an angle and its corresponding side plus the length of the corresponding side of the unknown angle. In other words, we must have ASA or SSA. The Law of Sines uses ratios of the length of a side to its corresponding angle. a b c sin A sin B sin C OR sin A sin B sin C a b c To calculate the measure of an angle or the length of a side, one ratio will be the ‘known’ pair and the other will consist of the other known side or angle. Example 5: Given ABC with A 42 0 , a = 52 cm, and C 73 0 cm, calculate the length of side ‘c’. Solution 5: a c sin A sin C 52 c 0 sin 42 sin 73 0 52 c .6691 .9563 .6691.9563 52 .6691.9563 c .6691 .9563 .956352 .6691c .956352 c .6691 74.3cm. c OR a c sin A sin C 52 c 0 sin 42 sin 73 0 52 sin 73 0 c sin 42 0 c 74.3cm. Solving the equation in terms of the variable ‘c’. Once the equation has been solved for ‘c’, the calculations can be done on the calculator. Not all students will be able to perform this task efficiently so the first method, although more cumbersome, may be the format that many students will follow. Example 6: Given ABC with B 39 0 , b = 5cm, and c 7.6cm. , calculate the measure of C (nearest tenth). Solution 6: b c sin B sin C 5 7 .6 0 sin C sin 39 7.6 sin 39 0 sin C 5 sin C .9565 sin 1 sin C sin 1 .9565 C 73.10 The answer given on the calculator for arcsin .9565 is 73.10. In the above example, the only measurement that was required was the measure of C . Therefore, two solutions for the angle had no bearing on the rest of the triangle. However, if we were solving the triangle, two solutions would have a direct impact on the triangle. To solve a triangle means to determine the measurements for all the sides and angles. Remember that the longest side of a triangle is opposite the largest angle. Ambiguous Case for the Law of Sines There is another possible answer to this question and that is the co-terminal angle of 106.90. The sine function is positive in the first and second quadrant, but calculators are designed to display the first angle as a result. To determine if the second angle is a possible solution, add 390 and 106.90. The sum is less than 1800, so both results are possible and acceptable solutions. If the sum of the two angles is greater than 1800, then the larger angle is not an acceptable value. This is referred to as the ambiguous case for the Law of Sines. Example 1: Given ABC with B 34 0 , b = 15cm, and c 20cm. , solve the triangle. Solution 1: b c sin B sin C 15 20 0 sin C sin 34 20 sin 34 0 sin C 15 sin C .7455 sin 1 sin C sin 1 .7455 C 48.2 0 or 131.8 0 If C = 48.20, then A = 97.80. This means that A is the largest angle of the triangle and side ‘a’ is the longest side. However, if C = 131.80, then A = 14.20. In this case A is the smallest angle of the triangle and then side ‘a’ is the shortest side. Therefore, the length of side ‘a’ must be calculated for each case. b a sin B sin A 15 a 0 sin 34 sin 97.8 0 15 sin 97.8 0 a sin 34 0 a 26.58cm. OR b a sin B sin A 15 a 0 sin 34 sin 14.2 0 15 sin 14.2 0 a sin 34 0 a 6.58cm. The two possible values for C creates two solutions for the measurements of the triangle when it is solved. The two solutions are shown below. A 97.8 0 A 14.2 0 B 34 0 B 34 0 C 48.2 0 a 26.58cm. OR C 131.8 0 a 6.58cm. b 15cm. b 15cm. c 20cm. c 20cm. Both solutions must be shown as the solution. Sometimes both the Law of Sines and the Law of Cosines can be used to determine the measure of an angle or the length of a side. Students will use the one that they are more proficient using. Example 2: Given ABC with A 46 0 , b 32cm. , and c 53cm. , find the measure of C (nearest tenth). Solution 2: Before the measure of C can be found, the length of side ‘a’ must be determined. a b 2 c 2 2bcCosA a 322 532 23253 cos46 0 a 1476.718807 a 38.43cm. Now that the length of side ‘a’ has been determined, the measure of C can be found. The lengths of the three sides are known so the Law of Cosines may be used or the Law of Sines may be used. The choice is simply an individual preference. a2 b2 c2 2ab 2 2 38.43 2 32 53 CosC 238.4332 308.1351 CosC 2459.52 CosC .1252 CosC cos 1 CosC cos 1 .1252 C 97.2 0 Exercises: 1. For each of the following triangles, determine the length of the indicated side or the measure of the indicated angle.(nearest tenth). a) Find ‘b’. b) Find B c) Find ‘a’. d) Find A Solutions: a) a b sin A sin B 17 b 0 sin 20.7 sin 118.3 0 17 sin 118.30 b sin 20.7 0 b 42.35 cm. b) a2 c2 b2 2ac 2 7 82 52 CosB 27 8 88 CosB 112 CosB .7857 CosB B 38.2 0 c) Find ‘a’. a b 2 c 2 2bcCosA a 6.7 2 8.42 26.7 8.4cos75.2 0 a 86.6970 a 9.31 cm. d) Find A a b sin A sin B 9.2 6.4 sin A sin 410 9.2 sin 410 sin A 6.4 sin A .9430 sin 1 sin A sin 1 .9430 A 70.6 0 The Law of Sines and the Law of Cosines are also used to solve real-world problems that can be represented by an oblique triangle. Example 1: An eight metre telephone pole has a very bad lean and creates an angle greater than 90 with the ground. A guide wire, 14 m long, is attached to the pole for support so the pole will not fall down. The guide wire is anchored in the ground at a point 10m from the base of the pole. Calculate the angle that the pole makes with the ground. Solution 1: a2 c2 b2 2ac 2 2 2 8 10 14 Cos 2810 32 Cos 160 Cos Cos 0.2 cos 1 Cos cos 1 0.2 101.5 0 The pole makes an angle of 101.50 with the ground. Example 2: A spider crawling down a wall spots its prey, a moth, on the ground at an angle of 160 with the wall. After crawling downward 16 cm, the moth still hasn’t moved, but now the angle with the wall is 280. How far is the moth from the wall? Solution 2: Before the distance CD can be calculated, some measurements must be determined. ACD 180 0 (90 0 28 0 ) ACB 74 0 62 0 ACD 62 0 ACB 12 0 BCD 180 0 90 0 16 0 BCD 74 0 The measurements of ABC will be used to calculate the length of ‘d’ ( AC ). ABC b c sin B sin C b 16 0 sin 16 sin 12 0 16 sin 16 0 b sin 12 0 b 21.21cm. ADC CD AC x sin 28 0 21.21 x .4694 21.21 sin CAD 21.21.4694 The moth is 9.96cm. from the wall. x 21.21 21.21 x 9.96cm Exercises: 1. Josh, Mary and Evan are playing frisbee in the school field. Their current positions form a triangle with the angle at Josh equal to 44o and the angle at Mary equal to 21o. If Mary is 15 metres from Evan, how far apart are Josh and Mary? 2. While exploring the woods at the end of Bengal Road in Mira, two of Glace Bay’s policemen, spotted a fire in the distance. From where they were standing, they estimated an angle of elevation of 150 to the top of the tower. Moving 10 m closer to the tower, they now estimate the angle of elevation to be 180. How high is the tower? Solutions: 1. The measure of E is 180 0 44 0 210 1150 e j sin E sin J e 15 0 sin 115 sin 44 0 15 sin 115 0 e sin 44 0 e 19.57 m The distance between Josh and Mary is 19.57m 2. Calculate the height of the tower BD. BCA 180 0 18 0 BCA 162 0 ABC 180 0 162 0 15 0 ABC 3 0 ABC a b sin A sin B a 10 0 sin 15 sin 30 10 sin 15 0 a sin 3 0 a 49.45m. BCD BD sin C BC x sin 18 0 49.45 x .3090 49.45 49.45.3090 x 49.45 49.45 x 15.28m The height of the tower is 15.28 m. Area of an Oblique Triangle The method that we will use to determine the area of an oblique triangle requires knowing the measurement of two sides of the triangle and the included angle (SAS). The formula used to calculate the area of ABC if b and c are the known sides and A is the included angle is: 1 Area bc sin A 2 Example 1: Determine the area of the following triangle: Solution 1: 1 ac sin B 2 1 Area 14.113.6 sin 410 2 Area 62.90cm 2 Area Example 2: Determine the area of the following triangle: The lengths of the three sides are known but the measure of an included angle must be determined. Choose one angle and use the Law of Cosines to find its measure. Solution 2: b2 c2 a2 cos A 2bc 8.42 8.12 4.82 cos A 28.4 8.1 cos A .8313 cos 1 cos A cos 1 .8313 1 bc sin A 2 1 Area 8.4 8.1sin 33.8 0 2 Area 18.93cm 2 Area A 33.8 0 If another angle were chosen, there may be a slight variation in the area of the triangle. This is the result of rounding values for the function(s) and any measurements that were calculated. Exercises: 1. Your backyard is a great spot to build a skating rink this winter. The ground must be covered with sand that costs $16 per square metre. If the sides of the triangular plot of land measure 8.24m., 7.67m. and 8.13m., what will be the cost of the sand needed to cover this area? 2. Josh, Mary and Evan are playing frisbee in the school field. Their current positions form a triangle with the angle at Josh equal to 44o and the angle at Mary equal to 21o. If Mary is 15 metres from Evan, what is the area of the triangular plot formed by Josh, Mary and Evan? Solutions: 1. b2 c2 a2 2bc 2 2 2 8.13 7.67 8.24 cos A 28.137.67 cos A .4572 cos A cos 1 cos A cos 1 .4572 A 62.8 0 1 bc sin A 2 1 Area 8.137.67 sin 62.8 0 2 Area 27.73m 2 Area The cost of the sand is (28m 2 )($16 / m 2 ) $448.00 Area 28m 2 2. In a previous exercise, we solved this problem to determine the distance between Mary and Josh. This distance was 19.57m. 1 ej sin M 2 1 Area 19.57 15 sin 210 2 Area 52.6m 2 Area