(1) Primary Trigonometric Ratios (SOH CAH TOA): Given a right triangle OPQ with acute angle, we have the following trig ratios: ADJ

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1 Tringles nd Trigonometry Prepred y: S diyy Hendrikson Nme: Dte: Suppose we were sked to solve the following tringles: Notie tht eh tringle hs missing informtion, whih inludes side lengths nd ngles. When sked to solve tringle, we must determine the mesure of ll of its ngles nd side lengths. In our ttempt to find these vlues, we re likely to onsider one or more of the following formuls. (1) Primry Trigonometri Rtios (SOH CAH TOA): Given right tringle OPQ with ute ngle, we hve the following trig rtios: sin = OPP HYP, ADJ os = nd tn = OPP HYP ADJ Here, HYP denotes the hypotenuse, whih is lwys ross from the right ngle. OPP denotes the side opposite/ross from the ngle ADJ denotes the side djent/eside the ngle, tht is not HYP. Applile to: right tringles only. To egin solving tringle, we need two known onstrints (e.g. two sides or one side nd n ute ngle) (2) Lw of Sines Given ny tringle ABC, we hve the following: sin A = sin B = sin C () sin A = sin B = sin C Applile to: ny tringle. To egin solving tringle, we need three known onstrints. Two of the three onstrints must inlude n ngle nd the side length ross from it. Mking Mth Possile 1 of 11 S diyy Hendrikson

2 Tringles nd Trigonometry Level: x 2 + y 2 =1 (3) Lw of Cosines Given ny tringle ABC, we hve the following: 2 = os C () 2 = os B () 2 = os A Applile to: ny tringle. To egin solving tringle, we need three known onstrints tht inlude either ll three side lengths or two side lengths nd the ngle in etween. Why re the ove formuls helpful when solving tringle? Answer: These formuls help us to relte the ngles of the tringle to its side lengths! (1) Primry Trig Rtios re only vlid when solving right tringles. (2) The Lws of Sines nd Cosines re vlid for ny tringle, so long s there is enough given informtion to sustitute into the formuls. (3) Although the Lws of Sines nd Cosines re pplile to right tringles, it is est prtie nd most e ient to use the Primry Trig Rtios. Below, we will explore these formuls nd pply them in some worked exmples! Given right tringle OPQ with ute ngle, we hve the following trig rtios: sin = OPP HYP, ADJ os = HYP nd tn = OPP ADJ Justifition: Notie tht the tringle in the unit irle hs the following side lengths: sin (height), os (se) nd 1 (sine the hypotenuse is rdius of the unit irle). Furthermore, this tringle is similr to 4OPQ y AAA similrity. Thus, y definition of similrity, we hve: sin 1 = OPP HYP nd os 1 = ADJ HYP ) tn = sin os = OPP HYP ADJ HYP = OPP ADJ Mking Mth Possile 2 of 11 S diyy Hendrikson

3 Tringles nd Trigonometry Level: x 2 + y 2 =1 Exerise 1: Solve the following tringles. Given ny tringle ABC, we hve the following: sin A = sin B = sin C () sin A = sin B = sin C Proof: Consider ny tringle ABC, oriented so tht the se is AC = (s shown) nd the ltitude, of length h, extends from the vertex t B. By our primry trig rtios, we know tht sin A = OPP HYP = h. Rerrnging, we hve tht h = sin A. It then follows tht the re of this tringle n e expressed s follows: Are = 1 2 se height = 1 2 sin A. We n lso orient the tringle so tht its se is AB = or BC =. From these orienttions, we find tht Are = 1 2 se height = 1 2 sin B or 1 2 sin C, respetively. In summry, we ve found tht: Are = 1 2 sin A = 1 2 sin B = 1 2 sin C Multiplying this eqution through y sin A = gives: = 1 2 sin B This is the required result, whih onludes the proof sin C ) sin A = sin B = sin C Mking Mth Possile 3 of 11 S diyy Hendrikson

4 Exploring Sine Lw Level: x 2 + y 2 =1 Consider the tringle ABC, s shown. In the digrm, \A is ross from side length, \B is ross from side length nd \C is ross from side length. This is onventionl nottion. We will ll the ngle nd the side length ross from it n opposites omo. The equtions quired from the Lw of Sines re s follows: sin A = sin B, sin A = sin C nd sin B = sin C Notie tht the ove equtions would only e helpful while solving tringle if we hve t lest three piees of informtion, whih n rise in one of the following wys: (1) Two ngles nd one side (resulting in n opposites omo nd n dditionl ngle), or (2) An opposites omo nd n dditionl side (resulting in two known sides). Notie tht if we re ever given tringle with two ngles, 1 nd 2, we n determine the third ngle y simply lulting the vlue of 3 = 180 ( ). Thus, ny side length tht is given will give us n opposites omo with one of the known ngles! For exmple, suppose we re given. Then, the opposites omo will inlude \B, whih we will lso know. Also, the given side length is n dditionl onstrint tht helps to produe unique tringle. Exerise 2: Solve the tringle elow, suh tht = 5, \A = 62 nd \C = 54. Mking Mth Possile 4 of 11 S diyy Hendrikson

5 Exploring Lw of Sines Level: x 2 + y 2 =1 Suppose we re given n ritrry tringle with onstrints onsisting of the following: two side lengths nd n ngle tht is ross from one of the given sides Question: Will these ontrints lwys produe unique tringle? Answer: Unfortuntely not. Any tringle with the onstrints desried ove will fll into one of the tegories given elow. The side ross from the given ngle will hve length tht is: (1) equl to the other known side length (2) shorter thn the other known side length (3) longer thn the other known side length In the se of (1), we hve n isoseles tringle, whih unovers n dditionl ngle. Therefore, we will hve unique tringle with ll known ngles nd one missing side, whih n e found esily. Figures 1-4 elow explore ll possile outomes tht fll under ses (2) nd (3). When onstruting digrm, note tht n opposites omo plus n dditionl side does not tell us nything out the ngle etween the sides whose lengths re given. In the figures elow, this would e \B. Tht mens tht when skething tringle with the given onditions, our sketh will e tenttive, sine it hnges when we vry the size of \B. These vritions re shown in the digrms. Some hoies of \B produe tringle tht stisfies the given onditions, while others do not! Figure 1 Figure 2 Figure 3 Figure 4 Figures 1, 2 nd 3 fll under se (2), while Figure 4 flls under se (3). Cse 1 Amiguous Cse ) Side ross from given ngle is shorter thn other known side Under this se, we hve three possile outomes. () No tringle exists with the given onstrints. (See Figure 1) This is euse the side length ross from the given ngle is too short! () Two tringles stisfy the given onstrints. (See Figure 2) () A right tringle stisfies the given onstrints. (See Figure 3) Mking Mth Possile 5 of 11 S diyy Hendrikson

6 Exploring Lw of Sines Level: x 2 + y 2 =1 Question: Cn we identify whih outome will our efore tully solving n exerise? Answer: Yes! However, given tht we must solve the tringle either wy, some folks prefer to dive into the exerise, wre tht ny of the three outomes my rise. If you re urious nd would like to know how determine the outome prior to solving, see the explntion elow. On the other hnd, if you would like to egin solving tringles nd disovering outomes s you go, see the Worked Exmples setion elow! Explntion Consider the following tringle, whih we sw during the proof for the Lw of Sines. Now, suppose tht we re sked to solve tringle with known vlues for \A nd side lengths nd, s seen in Figures 1-4. Rell tht our primry trig rtios gve tht the length of the ltitude is h = sin A, vlue tht we d e le to lulte using the given informtion. How does knowing the height of the tringle help us? Let s revisit Figures 1-3, elow. Eh figure hs distint reltionship with the height of the tringle. () No Solution (Figure 1): The height, h, isgreter thn the side ross from the given ngle. () Two Solutions (Figure 2): The height, h, isless thn the side ross from the given ngle. () A right tringle (Figure 3): The height, h, isequl to the side ross from the given ngle. If you find the ove results helpful, you re enourged to use them in the exerises tht follow! Exerise 3: Solve tringle ABC, with \B = 30, = 4 nd = 8. Mking Mth Possile 6 of 11 S diyy Hendrikson

7 Exploring Lw of Sines Level: x 2 + y 2 =1 Exerise 4: Solve tringle ABC, with \A = 31, = 62 nd = 130. Exerise 5: Solve tringle ABC, with \C = 27, = 12 nd = 24. Mking Mth Possile 7 of 11 S diyy Hendrikson

8 Exploring Lw of Sines Level: x 2 + y 2 =1 Cse 2 Unique Tringle ) Side ross from given ngle is longer thn other known side Rell Figure 4, shown elow. Here, the given onstrints determine unique tringle! Exerise 6: Solve tringle ABC, with \A = 61, = 5 nd = 2. Mking Mth Possile 8 of 11 S diyy Hendrikson

9 Exploring Lw of Cosines Level: x 2 + y 2 =1 Given ny tringle ABC, we hve the following: 2 = os C () 2 = os B () 2 = os A Proof: Consider 4ABC oriented in the oordinte plne, suh tht its se is AB =. Notie tht we n lwys onstrut the unit irle suh tht 4AED is similr to 4AF C. We know tht DE = sin A =sina (in Qudrnt II) nd AE = os A, dueto the onstrution in the unit irle. Thus, y the definition of similrity, we hve the following: CF = DE () CF = DE = sin A, nd 1 EF = AE () EF = AE = os A 1 We now hve the se nd height of 4AF C. Thus, the oordintes of C re ( os A, sin A). Note tht these oordintes would hve een the sme if \A ws right or ute. (Cn you verify this?) Next, we onsider the side length BC =. Applying the distne formul to the points B (, 0) nd C ( os A, sin A), we hve the following result: 2 =( os A) 2 +(0 sin A) 2 = 2 2 os A + 2 os 2 A + 2 sin 2 A = 2 2 os A + 2 (os 2 A +sin 2 A) = 2 2 os A + 2 (1) sine os 2 A +sin 2 A =1 = os A The result ove proves the third formul in the sttement of the lw. To quire the remining two, we simply orient the tringle in the oordinte plne suh tht the se is BC = or AC =, nd proeed with the sme resoning s ove. The Lw of Cosines is helpful when we re given three onstrints without n opposites omo. There re extly two instnes in whih this ours: (1) Given onstrints: three side lengths (2) Given onstrints: the lengths of two sides nd the ngle in etween Unlike with the Lw of Sines, these onstrints will lwys produe unique tringle! Next, we will ttempt some worked exmples! Mking Mth Possile 9 of 11 S diyy Hendrikson

10 Exploring Lw of Cosines Level: x 2 + y 2 =1 Exerise 7: Use the Lw of Cosines to determine the indited side, x. Exerise 8: Use the Lw of Cosines to determine the indited ngle,. Mking Mth Possile 10 of 11 S diyy Hendrikson

11 Bringing Things Together Level: x 2 + y 2 =1 Exerise 9: In the digrm elow, AB = 6, BD = 4, \ABD = 45 nd \CBD = 30. Determine the side length BC, indited y x. Mking Mth Possile 11 of 11 S diyy Hendrikson

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