Math 148 – Lab 4 – Trigonometry Winter Quarter, 2009 Due: Monday, March 9, 2009 Directions: • Work out each problem, and use the work to fill out the answer sheet. Sub- mit the answer sheet to your recitation instructor on or before the due date. • Material from this lab will appear on the final exam. • Late labs will not be accepted for grading. • When you are asked to “solve a triangle,” you must find the lengths of all three sides and the measures of all three angles of the triangle. In a triangle problem, you will need to decide which methods to employ to solve the triangle based on the conditions that are given. Purpose: • Insections9.1and9.2,youusedrighttriangletrigonometry(thePythagorean Theorem and the definitions of sine, cosine, and tangent) to solve right tri- angles. In section 9.3 you used the Law of Cosines and in section 9.4 you will use the Law of Sines to solve oblique triangles (triangles that are not right triangles). The Law of Cosines c2 = a2 +b2 −2abcosC The Law of Sines a sinA = b sinB = c sinC 1 Part I: Getting Ahead When the Law of Cosines does not help you solve an oblique triangle, what do you do? No, you don’t give up! Instead, seek another method. The next fact we use to solve oblique triangles is called the Law of Sines. The Law of Sines In triangle ABC, with the standard labeling of sides and angles as in the figure below, a sinA = b sinB = c sinC. C A B c b a Notice that the Law of Sines gives the equality of three ratios. Each ratio is the length of a side to the sine of the angle opposite that side. When working with the Law of Sines, we only use two of the three ratios at a time. The equation asinA = bsinB has four unknowns: two sides and two angles. The Law of Sines is useful when three of these quantities are given. It then follows that the Law of Sines can be used to solve triangles that meet either of the following conditions: 1. Two sides and the angle opposite one of the sides are given (SSA). 2. Two angles and a side are given (AAS). Readthroughthefollowingexamples, andthenworkouttheexercisesthatfolloweach example. We have only touched upon the Law of Sines. (You will learn more about the Law of Sines in lecture.) Inverse Sines Let θ be an angle in an oblique triangle. Then its sine will be a positive number which is less than 1. (If sinθ is equal to 1, then θ is a right angle.) Suppose x is positive and less than 1. Then the equationsinθ = x has acute-angled and obtuse-angled solutions. The acute angle is given by θ1 = sin−1 x. The obtuse angle is the supplement1 of the acute-angled solution and is given by: θ2 = 180◦ −sin−1 x. 1An angle and its supplement add up to a straight angle (180◦). So: supp θ = 180◦ −θ. 2 Example 1 (SSA) Solve triangle ABC under the conditions a = 30,b = 50,and B = 61◦. c o a = 30 b = 50 C B A 61 Solution: Because a,b,and B are given, we can use the equation asinA = bsinB to find A. Plugging in a = 30,b = 50,and B = 61◦, we find 30 sinA = 50 sin61◦ =⇒ sinA = 30sin61◦ 50 The SIN−1 key on the calculator shows that A ≈ 31.65288385◦. Store this result in your calculator under the letter ‘A.’ Rounding to the nearest tenth, A ≈ 31.7◦ Since we know A and B, we can compute C = 180◦ − A − 61◦ ≈ 87.34711615◦. Store this result in your calculator under the letter ‘C.’ Rounding to the nearest tenth, C ≈ 87.3◦ . Now we can use csinC = asinA or csinC = bsinB to find c. Let’s use the second one. Plugging in b = 50 and B = 61◦ and leaving C stored in the calculator, we find c sinC = 50 sin61◦ =⇒ c = 50sinC sin61◦ ≈ 57.1064353 Rounding to the nearest tenth, c ≈ 57.1 Caution: An SSA configuration may have no solutions, one solution, or two solutions. There is an obtuse angle A which solves the equation sinA = 30sin61 ◦ 50 , namely: A = 180◦ −31.65288385◦ = 148.34711615◦. But this obtuse angle does not give a second solution since any two angles of a triangle must add up to less than 180◦. Now you try: 1. Solve triangle ABC if b = 8, c = 9, and B = 51◦. (Round all results to the nearest tenth.) 3 Example 2 (AAS) Solve triangle ABC under the conditions A = 33◦,B = 46◦,and b = 4. c o 33 o B A C b = 4 a 46 Solution: Since A and B are given, we can compute C = 180◦ − 33◦ − 46◦ = 101◦. So C = 101◦ . We still must find a and c. To find a, we can use asinA = bsinB or asinA = csinC. Because c is not yet known, we’ll have to use the second equation. Plugging in A = 33◦,B = 46◦,and b = 4, we find a sin33◦ = 4 sin46◦ =⇒ a = 4sin33◦ sin46◦ ≈ 3.028549427 Rounding to the nearest tenth, a ≈ 3.0 . To find c, we have a choice of using csinC = asinA or csinC = bsinB. To avoid using the result from the previous computation, let’s use the second one. Plugging in C = 101◦,b = 4,and ,B = 46◦, we find c sin101◦ = 4 sin46◦ =⇒ a = 4sin101◦ sin46◦ ≈ 5.458489482 Rounding to the nearest tenth, c ≈ 5.5 . Now you try: 2. Solve triangle ABC if A = 66◦, C = 29◦, and c = 10. (Round all results to the nearest tenth.) 4 Part II: Which Method? Directions: Which method, Right Triangle Trigonometry, the Law of Cosines, or the Law of Sines, would be best to use as the first step in solving triangle ABC under the given conditions and why? Example 3 B = 65◦,a = 6,b = 7 Method: Law of Sines Why: We are given “SSA.” Example 4 a = 3,C = 90◦,c = 5 Method: Right Triangle Trigonometry (i.e. the Pythagorean Theorem and the definitions of sine, cosine, and tangent) Why: Triangle ABC is a right triangle Now you try: 3. Assume the angles of a triangle are labeled with the capital letters A,B, and C, and the side opposite an angle is labeled with the corresponding lowercase letter a,b, or c. Which method, “Right Triangle Trigonometry”, the “Law of Cosines”, or the “Law of Sines”, would be best to use as the first step in solving triangle ABC un- der the given conditions? To explain why, choose one of “right triangle”, “SSS”, “SAS”, “SSA”, or “AAS”. DO NOT SOLVE THE TRIANGLES. (a) A = 90◦, a = 5, b = 3. (d) a = 12, b = 16, c = 21.3. Which Method? Why? Which Method? Why? (b) C = 110◦, a = 16, b = 10. (e) A = 90◦, C = 60◦, b = 60. Which Method? Why? Which Method? Why? (c) A = 35◦, a = 7.5, b = 12. (f) B = 20◦, C = 31◦, b = 210. Which Method? Why? Which Method? Why? 5 Part III: Applications 4. Master Chief is trapped on an island with a Scorpion Tank as his only means of defense. An enemy Covenant jet fighter just landed on a smaller island across the water. Master Chief would like to fire a shell across the water directly at the jet. How- ever, astheScorpionTankcanonlylaunchashellamaximumhorizontaldistance of 10 miles, Master Chief may not be able to hit his target. He has decided to calculate the distance to the jet. For his calculation, he mea- sured five distances. Starting at a rock at point R, he drove his tank along a straight path in the direction of the jet, which is located at point J, to point A on the coast of the island. He placed at stake in the sand at point A and recorded RA = 1.9 miles. Next, from point A he drove 3.7 miles to a tree at point T. He then proceeded along a straight path from the tree at point T in the direction of the jet to point B at which he planted another stake into the sand. He determined that TB = 1.5 miles. In addition, he found BR = 3.9 miles and AB = 2.9 miles. Use the method below to find CJ, the distance between Master Chief and the Covenantfighterjet. (Rounddisplayedanswerstothenearesttenth.) WillMaster Chief be able to hit his target? Method: You can solve this problem by finding 7 measurements: (a) Find ∠RAB. (b) Which angle is supplementary to∠RAB? Find this angle. (c) Find ∠TBA. (d) Find the angle which is supplementary to∠TBA. (e) At this point, you should know two angles in triangle ABJ. Find the missing angle. (f) Find AJ. (g) Find CJ. Hints: You will need to use the following: • The Law of Cosines • Supplementary angles sum to 180◦ • The Law of Sines • The definition of a trigonometric function J C TR A B 5. A triangle has an angle φ whose sine is equal to 12. The angle is larger than 35◦ in measure. Find the exact measure (in degrees) of the angle φ. 6 Lab 4 Math 148 - Winter Quarter, 2009 Page 1 of 2 Answer Sheet – due Monday, March 9, 2009 Name: Recitation Time: 1. Solve triangle ABC if b = 8, c = 9, and B = 51◦. Number of Solutions (0, 1 or 2): angle A angle C side a First solution: Second solution: 2. Solve triangle ABC if A = 66◦, C = 29◦, and c = 10. angle B side a side b Exactly one solution: 3. (a) A = 90◦, a = 5, b = 3. Which method? Why? (b) C = 110◦, a = 16, b = 10. Which method? Why? (c) A = 35◦, a = 7.5, b = 12. Which method? Why? (d) a = 12, b = 16, c = 21.3. Which method? Why? (e) A = 90◦, C = 60◦, b = 60. Which method? Why? (f) B = 20◦, C = 31◦, b = 210. Which method? Why? 7 Name: Page 2 of 2 Round all displayed answers to the nearest tenth. (But be sure to use all available precision when using a value in a calculation.) 4. (a) ∠RAB= . (b) Which angle is supplementary to∠RAB? . What is the measure of this angle? . (c) ∠TBA= . (d) Which angle is supplementary to∠TBA? . What is the measure of this angle? . (e) At this point, you should know two of the three angles in triangle ABJ. What is the measure of the third angle? . (f) AJ= . (g) Find CJ= . Will Master Chief be able to hit his target? (Circle your answer.) Yes No 5. φ= . 8