CHAPTER 6: ADDITIONAL TOPICS IN TRIG

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1 (Setion 6.1: The Lw of Sines) 6.01 CHAPTER 6: ADDITIONAL TOPICS IN TRIG SECTION 6.1: THE LAW OF SINES PART A: THE SETUP AND THE LAW The Lw of Sines nd the Lw of Cosines will llow us to nlyze nd solve oblique (i.e., non-right) tringles, s well s the right tringles we hve been used to deling with. Here is n exmple of onventionl setup for tringle: There re 6 prts: 3 ngles nd 3 sides. Observe tht Side fes Angle A, b fes B, nd fes C. (In right tringle, C ws typilly the right ngle.) When we refer to, we my be referring to the line segment BC or its length. The Lw of Sines For suh tringle: sin A = Equivlently: b sin B = sinc sin A = sin B b = sinc

2 PART B: WHAT MUST BE TRUE OF ALL TRIANGLES? (Setion 6.1: The Lw of Sines) 6.02 We ssume tht A, B, nd C re ngles whose degree mesures re stritly between 0 nd 180. The 180 Rule The sum of the interior ngles of tringle must be 180. Tht is, A + B + C = 180. The Tringle Inequlity The sum of ny two sides (i.e., side lengths) of tringle must exeed the third. Tht is, + b >, b + >, nd + > b. Think: Detours. Any detour in the plne from point A to point B, for exmple, must be longer thn the stright route from A to B. Exmple: There n be no tringle with side lengths 3 m, 4 m, nd 10 m, beuse 3+ 4 >10. If you hd three pik-up stiks with those lengths, you ould not form tringle with them if you were only llowed to onnet them t their endpoints. The Eting Rule For given tringle, lrger ngles fe (or et ) longer sides. You n use this to hek to see if your nswers re sensible.

3 (Setion 6.1: The Lw of Sines) 6.03 PART C: EXAMPLE Exmple Given: B = 40, C = 75, b = 23 ft. Solve the tringle. In your finl nswers, round off lengths to the nerest foot. Solution Sketh model tringle. (Informtion yet to be determined is in red.) Find Angle A: Use the 180 Rule. A = A = 65 o Use the Lw of Sines: sin A = b sin B = sin65 = 23 sin 40 = sinc sin75

4 Find : (Setion 6.1: The Lw of Sines) 6.04 Observe tht the middle rtio is known, so we should use it when we solve for both nd. Solve sin65 = 23 for. sin 40 Find : = 23sin65 sin feet Wrning: Mke sure your lultor is in DEGREE mode. Wrning: Avoid pproximtions for trig vlues in intermedite steps. Exessive rounding n render your finl pproximtions inurte. Memory buttons my help. If fesible, you should keep ext expressions suh s sin65 throughout your solution until the end. Wrning: Don t forget units where they re pproprite! Solve 23 sin 40 = for. sin75 23sin75 sin 40 = 35 feet Wrning: Although you ould, in priniple, use the sin65 rtio 23 insted of the rtio, it is ill-dvised to use your rough sin 40 pproximtion for s foundtion for your new lultions.

5 (Setion 6.1: The Lw of Sines) 6.05 Use the Eting Rule to Chek: Mke sure tht lrger ngles et longer sides. PART D: CASES If you re given 3 prts of tringle (inluding one side), you n solve the tringle by finding the other 3 prts or by disovering tht there is no tringle tht supports the given onfigurtion (in whih se there is no solution ). Two tringles re onsidered to be the sme if they hve the sme vlues for A, B, C,, b, nd. If you re only given the 3 ngles (the AAA se), then you hve n entire fmily of similr tringles of vrying sizes tht hve those ngles. On the other hnd, it is possible tht no tringle n support the given onfigurtion. For exmple, mybe the Tringle Inequlity is being violted. In the Ambiguous SSA se, whih is disussed in Prt E, two different tringles my work. The Lw of Sines is pplied in ses where you know two ngles nd one side. For exmple, it is pplied in: The AAS se (suh s in the previous Exmple, in whih we re given Angle B, Angle C, nd Side b, noninluded side), nd The ASA se (in whih, for exmple, we re given Angle A, Side b, nd Angle C; here, Side b is inluded between the two given ngles). Cn you see how these ses n yield either no tringle or extly one? The Lw of Sines is lso pplied in the Ambiguous SSA se, desribed next.

6 (Setion 6.1: The Lw of Sines) 6.06 PART E: THE AMBIGUOUS SSA CASE The SSA se is when you re given two sides nd noninluded ngle. For exmple, you ould be given Side, Side, nd Angle C (in purple below). The SSA se is lled the mbiguous se, beuse two tringles (tht is, two tringles tht re not ongruent) my rise from the given informtion. Ber in mind tht the possibility of no tringles potentilly plgues ll ses. In the figure bove, imgine the side lbeled being rotted bout Point B. We n obtin seond tringle (in blue dshed lines below) for whih the given informtion still holds! How is this issue refleted in the Lw of Sines? Look t the Lw of Sines: sin A = b sin B = sinc In order to solve the tringle, we find Angle A using the Lw of Sines. We ould then use the 180 Rule to find the remining ngle, Angle B. The problem is tht we must first find sin A, nd n ute ngle nd its supplementry obtuse ngle my shre the sme sin vlue. (Look t the Unit Cirle!) These two possibilities my yield two different tringles, provided tht the 180 Rule does not fll prt (the obtuse ngle my et up too mny degrees). Also, it s gme over if sin A ws not in 0,1 ( to begin with.

7 (Setion 6.1: The Lw of Sines) 6.07 PART F: THE AREA OF A TRIANGLE In the SAS Cse (desribed below), you n quikly ompute the re of tringle. Are of Tringle (SAS Cse) Let nd b be two sides (i.e., two side lengths) of tringle, nd let C be the inluded ngle between them. Then, the re of the tringle is given by: Are = 1 2 bsinc Think: Hlf the produt of two sides nd the sine of the inluded ngle between them (represented by different letter). Wrning: Avoid writing A for Are, sine we often use A to nme vertex on the tringle. Wht hppens if C is right ngle? Vritions The following lso hold: Are = 1 2 bsin A Are = 1 2 sin B If you know one of the re formuls bove, you n figure out the other two. If you re given word problem or n unlbeled tringle, you ould ssign lbels in mnner best suited for the formul you re most fmilir with.

8 (Setion 6.1: The Lw of Sines) 6.08 Proof Without loss of generlity, let s sy we re given, b, nd C in the figure below. The h represents the height of the tringle, provided tht b is tken s the bse. Sine sinc = h, the height h of the tringle is given by: h = sinc. The re is then given by: Are = 1 ( 2 bse )( height) = 1 2 bh = 1 2 b ( sinc ) = 1 2 bsinc This proof is similr to the proof for the Lw of Sines on p.468 of Lrson, in whih expressions for the height of tringle involving the sines of the three ngles re equted.

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