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# Lecture 20, 21, and 22 Math 240

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This page Class Notes was uploaded by AnnMarie on Wednesday November 4, 2015. The Class Notes belongs to Math 240 at Louisiana Tech University taught by Jonathan B Walters in Fall 2015. Since its upload, it has received 64 views. For similar materials see Precalculus in Mathematics (M) at Louisiana Tech University.

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Date Created: 11/04/15

Chapter 63 Trigonometric Functions of Real Numbers This section is mostly review of finding Trigonometric functions using reference angles Review of the Reciprocal Identies and Pythagorean Identies However finding the area of a triangle is introduced within this section In class we started with an example which was review of using a reference angle to evaluate a trigonometric function 17H Example Find a seC 3 and b CSCL3H Solution a The angle 13135 coterminal with and these angles are in Quadrant IV Thus the reference angle is 211 131 2 Since sec mthe value of seal is positive S C1I S C53 H wig 2 b Using the same reference angle as in part a we find that 17H CSC 3 1 2 sin 2 3 Area of a Triangle The area A of a triangle with sides of lengths a and b and with included angle is A ab sin Example Find the area of the shaded region with r 12 and H e g Solution A Asector A0 A r2 12r25in E 3144 4sin 72 24H 2 2411 363 This would be an accepted answer on the test 13 units2 22 Suggested Assignments for Chapter 63 512 Find the refenced angle for the given angle 5 a 1200 b 2000 c 2850 7 a 2250 b 8100 c 1050 9 a b 9 c 107E 11 a b 14n c 14 1336 Find the exact value of the Trigonometric Function 13 cos 1500 15 tan 3300 17 cot1200 19 csc6300 21 cos 5700 23 tan 7500 25 sin 37 27 tan 29 csc 31 sec 33 cot 35 tan 3740 Find the quadrant in which H lies from the given information 37 sin H lt Oand cos H lt O 39 sec H gt Oand tan H lt 0 4754 Find the balues of the Trigonometric Functions of H from the information given 47 sin H 5 H in Quadrant IV 49cos sin H lt0 51 csc H 2 H in QuadrantI 53 cos H tan H lt O 55 If H g find the value of sin2 ancl 25in 57 Find the area of the triangle with sides of length 7 and 9 and included angle 72 Chapter 65 Law of Sines If you recall from Section 62 we were able to solve right triangles with Trigonometric Rations However not all triangles are right triangles In the figure below there are four triangles which three of them have for one or more sides These triangles represent four different cases that you may come across when asked to find the missing sides and angles of the triangle The first two cases we can use the Law of Sines to solve and the last two cases can be solved using the Law of Cosine In this Section we will learn how to solve for nonright triangles using the Law of Sines And later on we will learn how to solve triangles using the Law of Cosines SA or Case 1 One side and two angles are given Case 2 Two sides and the opposite angle of one of the sides is given Case 3 Two sides and one of the angles are given Case 4 All sides are given However since we are using the lecture notes from class we will start with an example that will review how to solve a right triangle Example Find the remaining angle and side Solution 180 15 90 75 sin150 4sin150X X7 What if the triangle is not a right triangle You will use the law of sines to solve Laws of Sines In triangle ABC we have smA sinB san a b C 0quot 0 b C Example A satellite orbiting the earth passes directly overhead at observation stations that are 340 miles apart At an instant when the satellite is between the two stations its angle of elevation is simultaneously observed to be 60 and 75 How far is the satellite from each observatory Solution We know that c is 340 miles because it runs between the two observations The angle of A is 75 and the angle of B is 60 To find angle C we will subtract 75 and 60 from 180 C 1800 75 60 C 45 Now we can use the law of sines to find lengths a and b a C 3405in75 sinA sin C 2 a sin45 a 46445 miles L L sinB sinC b sin45 b 41641 miles Example Solve the triangle Solution c 1800 3750 2810 c 11440 sin B sin A 51112810 sin375 17sin 281 b sin375 b W 1315 17 sirfC sinLA Z sin1cl44 sin375 17sin1442 sin375 C W 2543 When solving for SSA cases there are three possibilities 1 Unique Triangle 2 Two Triangles satisfy information 3 No Triangles satisfy the information The last three examples will show how determine which possibility works for the given problem Example Determine how many triangles satify A 4310 a 1862 and b2486 Solution smA sinB 2 Sin 4310 sinB a b 1862 2486 2486 sin431 1862 Sin B 0912255 sinB Since sin B 0912255 can be found in two quadrants then there are two triangles B1 is in Q1 B1 sin3910912255 65820 B2 is in Q11 32 180065820 114180 AAlBlcl c1 1800431065820 c1 71080 1862 sin7108 C1 sin431 c1 25778 AAZBZCZ c2 18004310114180 c1 22720 C 1862 sin2272 1 sin431 c1 10525 Example Determine how many triangles satify A 1100 a 28 and b15 Solution sigB 517 gt 28 sin B 15 sin110 15sin110 28 sin B m 0503 sinB Since sin B W 0503 can be found in two quadrants then there are two triangles B1 is in Q1 Bl sin3910503 30230 B2 is in Q11 32 180030230 149770 AAlBlcl c1 18003o2301100 c1 3977O 285in3977 C1 sin110 c1 1906 AAZBZCZ C2 18001100149770 C1 70 0 There is no second triangle so there is only one triangle Example Determine how many triangles satify A 420 a 42 and b142 Solution sigB siZA 2 Sin B 1425213420 sinB 2262 1 lt sin x S 1for allX thus there are no triangles Chapter 65 Suggested Assignments 3 8 Use the Law of Sines to find the indicated side X or angle 9 1318 Sketch each triangle and then solve the triangle using the Law of Sines 13 A 500 B 680 and c 230 15 A 300 C 650 and b 10 17 B 290 C 510 and b 44 1928 Use the Law of Sines to solve for all possible triangles that satisfy the given conditions 19 a 28 b 15 and A 1100 21 a 20 c 45 and A 1250 23 b25c30and 8250 25 a 50 b 100 and A 500 27 a 26 c 15 and C 290 29 For the triangle shown find a angle BCD and b angle DCA fl 33 To find the distance across a river a surveyor chooses points A and B which are 200 ft apart on one side of the river see figure She then chooses a refernce point C on the opposite side of the river and finds that angle BAC is about 82 degrees and angle ABC is about 52 degrees Approximate the distance from A and C u j q l 15 C V 39 f EA WLaw 35 mgah GNm A ABC 06 Caz A b2 e k gitv Ra CW 3 fa ak Wm lash 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