Unit 6 Solving Oblique Triangles - Classwork

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1 Unit 6 Solving Oblique Tringles - Clsswork A. The Lw of Sines ASA nd AAS In geometry, we lerned to prove congruence of tringles tht is when two tringles re exctly the sme. We used severl rules to prove congruence: Angle-Side-Angle (ASA), Angle-Angle-Side (AAS), Side-Angle-Side (SAS) nd Side-Side-Side (SSS). In trigonometry, we tke it step further. For instnce, if we know the vlues of two ngles nd side of tringle, we cn solve tht tringle tht is we cn find the other ngle nd the other sides. We hve lerned to solve right tringles in Unit 3. In this section we lern how to solve oblique tringles tringles tht do not hve right ngle. First, let s strt with generliztion for this section. All tringles will hve 6 pieces of informtion 3 ngles nd 3 sides. The ngles re lbeled A, B, nd C nd the sides re opposite the ngles nd re lbeled, b, nd c. Note tht right ngle trigonometry doesn t help us here. There is no right ngle, thus no hypotenuse. We need something else. Tht something is clled the Lw of Sines. In both tringles, we drop perpendiculr h. In both tringles we cn then sy : The Lw of Sines If ABC is tringle with sides, b, nd c, then The Lw of Sines is relly three lws in one: sin A = b sin B, b sin B = re 4 vribles. If you know 3 of them, you cn find the 4 th. sin A = h b nd sin B = h So h = bsin A nd h = sin B It follows tht sin B = bsin A nd thus sin A = c sinc nd sin A = b sin B b sin B = sin A = c sinc c. In ech one, there sinc Exmple 1) AAS A = 23 =14 B = 48 b = C =109 c = Exmple 2) - ASA A = 72.4 B = 68.9 C = 38.7 = 3.88 in b = 3.8 in c = 2.55 in Most work is done on the clcultor. Your job is to show wht formuls you re using. To check problem, verify tht the lrgest ngle is opposite the lrgest side nd the smllest ngle is opposite the smllest side. 6. Oblique Tringles Stu Schwrtz

2 B. The Lw of Sines - SSA One of the rules for congruence is not Side Side- Angle (SSA). You my hve wondered why not. The problem with SSA is tht while 2 sides nd n ngle my identify tringle, it is possible tht the tringle my not exist with tht informtion. Or it is possible tht 2 sides nd n ngle my identify two possible tringles. Let s exmine the SSA cse more in depth nd find tht it breks up into 6 possible situtions. For ll 6 situtions, we will ssume tht you re given, b, nd A. Before we ctully ttempt to solve tringle in the SSA cse, we must decide which of the 6 situtions bove the problem form is. In most cses, simple drwing cn help us decide. Exmple 3) Exmple 4) A = 62 = 5 B = b =15 A = 31 = 22 B = b = 9 Tringles Possible: None Tringles possible: One 6. Oblique Tringles Stu Schwrtz

3 Exmple 5) Exmple 6) A = 99 = 9.2 B = b = 5.5 A =125 =16 B = b = 30 Tringles Possible: One Tringles possible: None Exmple 7) Exmple 8) A = 30 = 7 B = b =14 A = 25 =11 B = b =14 Tringles Possible: One Tringles possible: One Exmple 9) Exmple 10) A = 38 =12.9 B = b = 21 A = 38 =13 B = b = 21 Tringles Possible: None Tringles possible: Two 6. Oblique Tringles Stu Schwrtz

4 In the lst two exmples, we find tht exmple 9 is not possible to drw while exmple 10 is not only possible, but two tringles re possible (it is clled the mbiguous cse). Yet there is only 0.1 difference in the vlue of. It is impossible to tell by eye. So how cn you tell? The nswer lies in the Lw of Sines. If we set up the Lw of Sines with the A nd B fmily, we get sin B = bsin A nd sin B = bsin A. Since we know tht the lrgest vlue of sin B is 1, we cn determine whether tringle cn exist. If sin B = bsin A >1, the tringle is impossible. if sin B = bsin A <1, there re two tringles possible. So let us show tht Exmple 9 hs no tringle possible while tringle 10 hs two tringles possible. Exmple 9) - clcultions A = 38 =12.9 B = b = 21 sinb = bsin A = Exmple 10) - clcultions A = 38 =13 B = b = 21 sinb = bsin A = Note tht the only time we need to go through this step is when it is uncler from the drwing whether the tringle cn be drwn. This occurs when there is question when the SSA problem is sitution 1 or sitution 3 (sitution is very rre s it yield perfect right tringle.) Another wy to determine quickly if obtuse tringles re possible is to remember the fct tht in ny tringle, the lrgest ngle must lwys be opposite the lrgest side. Exmple 6 bove clerly is impossible to drw becuse ngle A =125 must be the lrgest ngle in the tringle. Which mens tht side must be the lrgest side. But since is given to be16 nd b is given to be 30, the tringle is impossible. IMPORTANT: Also note tht this phenomenon only occurs in n SSA sitution. ASA nd AAS re lwys defined with one nd only one tringle nd the Lw of Sines solves this esily. Now tht we hve decided how to determine whether or not n SSA problem hs solution, we need to ctully solve it. Agin, let s ssume tht we re given, b, nd ngle A nd the SSA problem hs only one solution. # bsin A& Step 1: Find ngle B by using the fct tht B = sin "1 % ( $ ' Step 2: Find ngle C by using the fct tht A + B + C =180 so C =180 " A " B Step 3: Use the Lw of Sines with the A fmily nd the C fmily. sin A = c sinc so c = sinc sin A Step 4: Quickly verify your nswer by checking to see the lrgest side is opposite the lrgest ngle nd the smllest side is opposite the smllest ngle. Let s do severl problems tht we exmined before. 6. Oblique Tringles Stu Schwrtz

5 Exmple 11) Exmple 12) A = 31 = 22 B =12.16 b = 9 C = c = A = 99 = 9.2 B = b = 5.5 C = c = 6.56 Exmple 13) Exmple 14) A =13.42 =19 B =160 b = 28 C = 6.58 c = 9.38 A = " =1 ft 10 in B = " b = 2 ft in C = " c = 2 ft 2 in Finlly, let s go through the procedure we use when you determine tht there re two solutions. Remember, the only wy this cn hppen is if A < 90 nd < b. You drw the sitution nd relize tht there re two wys to drw the tringle. When in doubt use the fct tht if sin B = bsin A if sin B = bsin A <1, there re two tringles possible. Assuming tht there re two tringles possible, here is the procedure. 1. Drw the sitution. Show both tringles. "ABC will be the cute tringle nd "A B # C will be the obtuse tringle. 2. Solve "ABC first using the procedure bove. ( reminder) # bsin A&. Find B by using the fct tht B = sin "1 % ( $ ' b. Find C by using the fct tht C =180 " A " B c. Find big c by using the fct tht c = sinc sin A 3. Solve "A B # C by using the procedure below.. Find ngle B in "B B # C by relizing tht "B B # C is isosceles. b. Find obtuse ngle B in "A B # C by using the fct tht "A B # C =180 $"B B # C c. Find little ngle C by using the fct tht C =180 " A " B # d. Find little c by using the fct tht c = sinc sin A >1, the tringle is impossible nd 6. Oblique Tringles Stu Schwrtz

6 This looks like lot of work but tringle ABC is solved using the bsic procedure for one tringle. The fct tht tringle BB C is isosceles mkes solving the obtuse tringle esy. Exmple 13) Exmple 14) A = 25 =11 B = b =14 C = c = A = 38 =13 B = 84 b = 21 C = 58 c =17.91 A = 25 =11 B " = b =14 C = 7.54 c = 3.42 A = 38 =13 B " = 96 b = 21 C = 46 c =15.19 We hve now solved tringles in the form of ASA, AAS, nd if possible, SSA. Tht leves us with the cses of SAS, nd SSS. Here re two exmples. Explin why the Lw of Sines does not help us solve them. Exmple 15) Exmple 16) A = 68 = B = b = 8 c = 5 A = = 4 B = b = 7 c = 9 C. The Lw of Cosines Since the Lw of Sines is only helpful when we hve complete fmily ( fmily or b fmily, or c fmily) nd hlf of fmily, we need nother lw. Tht lw is the Lw of Cosines. In ny tringle with sides, b,nd c nd ngles A, B, nd C. y 2 + k 2 = 2 ( ) 2 + y 2 ( ) 2 + ( bsin A) 2 2 = c " x 2 = c " bcos A 2 = c 2 " 2bccos A + b 2 cos 2 A + b 2 sin 2 A ( ) 2 = c 2 " 2bccos A + b 2 cos 2 A + sin 2 A 2 = b 2 + c 2 " 2bccos A b 2 = 2 + c 2 " 2ccosB c 2 = 2 + b 2 " 2bcosC 6. Oblique Tringles Stu Schwrtz

7 Notice tht like the Lw of Sines, the Lw of Cosines is relly three lws. In SAS problems, you lwys use the one for which the ngle is given. So here is the technique to solve problems in the form of SAS. Note tht no drwing is relly necessry. Step 1) Use the Lw of Cosines for which you re given the ngle. For instnce, if you re given ngle A, you will use 2 = b 2 + c 2 " 2bccos A which llows you to sy tht = b 2 + c 2 " 2bccos A. Input directly into your clcultor. Step 2) Now tht we hve complete fmily, we now switch to the Lw of Sines. Only use the Lw of Cosines once in problem. IMPORTANT: to void possible problem, lwys use the Lw of Sines to find the smllest ngle remining. So use the Lw of Sines with your complete fmily nd the smllest side remining. Step 3) You hve two ngles. Find the other ngle by subtrcting the sum of your two known ngles from 180. Step 4) As before, verify your nswers by checking tht the lrgest ngle is opposite the lrgest side nd the smllest ngle is opposite the smllest side. Exmple 17) Exmple 18) A = 68 = 7.68 B = b = 8 C = c = 5 A = 4.42 = 6 B =14.3 b =19.24 C = c = 25 The technique for solving problems in the form of SSS is similr in tht we hve to use the Lw of Cosines. But since we hve 3 sides nd no ngles, we must solve for n Angle first. Step 1) You cn use ny of the 3 Lws of Cosines, However to void potentil problem, use the Lw of Cosines to find the lrgest ngle. If is the lrgest side, use 2 = b 2 + c 2 " 2bccos A which llows you to sy tht cos A = b2 + c 2 " 2. 2bc Clculte the vlue of the right side of this eqution on your clcultor nd tke the inverse cosine of tht nswer to find A. Step 2) Now tht we hve complete fmily, we now switch to the Lw of Sines. Only use the Lw of Cosines once in problem. So use the Lw of Sines with your complete fmily nd ny other hlf fmily. Step 3) You hve two ngles. Find the other ngle by subtrcting the sum of your two known ngles from 180. Step 4) As before, verify your nswers by checking tht the lrgest ngle is opposite the lrgest side nd the smllest ngle is opposite the smllest side. Exmple 17) Exmple 18) A = = 7 B = b = 9 C = c =12 A = = 6000 B = b = 4000 C = c = Oblique Tringles Stu Schwrtz

8 D. Are of Oblique Tringles There re two formuls using trigonometry tht will llow us to find the re of oblique tringles bsed on given informtion. Obviously, if the tringle is right tringle, we only need both legs: Are = 1 ( 2 bse )( height). Are = 1 ( 2 bse )( height) Are = 1 2 cbsin A So we sy Are = 1 2 bcsin A Similrly, Are = 1 2 csin B Similrly, Are = 1 2 bsinc This formul works when you hve two sides nd the included ngle (SAS). But frequently you hve three sides of tringle nd wish to determine the re. In tht cse, we hve nother formul tht will determine the re of tht tringle. It is clled Heron s (pronounced Hero s) formul. Heron's Formul If "ABC hs sides,b, nd c, the re of the tringle is given by Are = s( s # ) ( s # b) ( s # c) where s = + b + c 2 Find the res of the following tringles: Exmple 19) Exmple 20) A = 68 = B = b = 8 c = 5 A = = 6 B =14.3 b = c = 25 Are = Are = Exmple 21) Exmple 22) A = = 7 B = b = 9 c =12 A = = 6000 B = b = 4000 c = 5000 Are = Are = 9, Oblique Tringles Stu Schwrtz

9 Exmple 23) Exmple 24) A = 31 = 22 B =12.16 b = 9 A = 25 =11 B = b =14 C = C = Are = Are = E. Applictions of Solving Oblique Tringles Finlly, Now tht we cn solve oblique tringles nd find their re, we turn to pplictions of these topics. When solving pplictions, lwys following these procedures. 1) Drw picture describing the sitution. 2) Lbel the picture with vribles nd be sure you use these vribles in your solution. 3) Identify whether the resulting tringle is n ASA, AAS, or SSA sitution requiring the Lw of Sines or n SAS or SSS requiring the Lw of Cosines. Sometimes you will hve to use these lws first nd then use right ngle trigonometry. " 4) If n re is required decide whether the SAS formul $ Are = 1 # 2 bcsin A % ' is & needed or Heron s formul where SSS is required. 5) Alwys remember to nswer the question(s) sked. 6) Alwys remember to supply proper units to the nswer(s). Circle the nswer(s). Exmple 25) A tree lens 7 to the verticl. At point 40 feet from the tree (on the side closest to the len), the ngle of elevtion to the top of the tree is 24. Find the height of the tree. = 40sin24 sin73 =17.01 ft Exmple 26) Two mrkers A nd B re on the sme side of river re 58 feet prt. A third mrker is locted cross the river t point C. A surveyor determines tht "CAB = 68 nd "ABC = 52. ) Wht is the distnce between points A nd C? b) Wht is the distnce cross the river? b = 58sin52 = ft sin60 d = 52.78cos22 = ft 6. Oblique Tringles Stu Schwrtz

10 Exmple 27) A hot ir blloon is hovering over Vlley Forge. Person A views the blloon t n ngle of elevtion of 25 while person B views the blloon t n ngle of elevtion of 40. If A nd B re 4000 feet prt, find the height of the blloon. = 4000sin25 = ft sin115 h = sin 40 = 1, ft Exmple 28) Do the problem bove with the sme ngles nd distnces but ssume tht the people re now on the sme side of the blloon s seen in the ccompnying picture. = 4000sin25 = ft sin15 h = sin 40 = 4, ft Exmple 29) Two torndo spotters re on rod running est-west nd re 25 miles prt. The west mn spots torndo t bering N 37 E nd the est mn spots the sme torndo on bering of N 56 W. How fr is the torndo from ech mn nd how fr is the torndo from the rod? = 25sin53 = miles sin103 b = 25sin24 =10.44 miles sin103 d = 20.49sin53 = miles 6. Oblique Tringles Stu Schwrtz

11 Exmple 30) A smll ship trvels from lighthouse whose light only shines est on bering of N 70 E distnce of 8 miles. It then mkes turn to the right nd trvels for 5 miles nd finds itself directly in the bem of the lighthouse s light. How fr from the lighthouse is the ship? c = 5sin = miles or sin20 c = 5sin13.18 = 3.33 miles sin20 Exmple 31) On regultion bsebll field, the four bses form squre whose sides re 90 feet long. The center of the pitching mound is 60.5 feet from home plte. How fr is the mound from first bse? d = " 2( 60.5) ( 90)cos45 d = ft Exmple 32) To determine the distnce cross lke AB, surveyor goes to point C where he cn mesure the distnce from C to A nd from C to B s well s the ngle ACB. If AB is 865 feet nd BC is 188 feet nd "ACB = #, find the distnce AB cross the lke. d = " 2( 865) ( 188)cos42 18 # d = ft Exmple 33) Circulr trcts of lnd with dimeters 900 meters, 700 meters nd 600 meters re tngent to ech other externlly. There re houses directly in the center of ech circle. Wht re the ngles of the tringle connecting the houses nd wht is the re of tht tringle? A = = 800 B = b = 750 C = c = 650 # " & # 650sin69.28 & A = cos "1 % ( C = sin "1 % ( $ 2( 750) ( 650) ' $ 800 ' Are = 1100( 300) ( 350) ( 450) = 68, meters 2 6. Oblique Tringles Stu Schwrtz

12 Unit 6 Solving Oblique Tringles - Homework 1. Solve ech of the following tringles. 3 deciml plces nd deciml degrees unless DMS given. A = 62 = = 9.93 A = =19.4. B = 78 b =11 b. B = 72.4 b =19.82 C = 40 c = c = 7.23 C = 38.7 c =13.00 A =19.2 = A =122.5 = 5.87 ft c. B = b = d. B = 38.2 b = 4.30 ft C = 8.7 c =12.4 C =19.30 c = 2.3 ft A = " =1.01 A = " =1.61 mm e. B = " b = 2.97 f. B = " b = 3.84 mm C = " c = 2.56 C = " c = 3.69 mm 2. Determine how mny tringles re possible in the following SSA situtions. Do not ctully solve the tringles. A =152 = 42 A = 31 = 6. B = b = 55 b. B = b = 4.5 Tringles Possible: None Tringles possible: One 6. Oblique Tringles Stu Schwrtz

13 A =151 =125 A = 68 =12 c. B = b = 79 d. B = b = 55 Tringles Possible: One Tringles possible: None A = 27 =12 A = 58 = 25 e. B = b =17 f. B = b = 30 Tringles Possible: Two Tringles possible: None A = = 40 A = = g. B =122 b = 38 h. B =142 b =19 c = 6 Don t get upset tht you ren t given A,, nd b in exercises g - j. You cn still drw the pictures. Tringles Possible: None Tringles possible: One 6. Oblique Tringles Stu Schwrtz

14 A = = A = = i. B = b = 6 j. B = b = 50 C = " c =16 C = " c = 38 Tringles Possible: One Tringles possible: Two 3. These re some of the sme problems in section 2 bove those tht hve 1 or 2 solutions. Solve them. If they hve two solutions, you will need to mke nother chrt. A = 31 = 6 A =151 =125. B = b = 4.5 b. B =17.84 b = 79 C = c = 9.39 C =11.16 c = A = =13.91 A = " = c. B =142 b =19 d. B = " b = 6 C =11.21 c = 6 C = " c =16 6. Oblique Tringles Stu Schwrtz

15 A = 27 =12 A = " = 3.31 ft e. B = b =17 f. B = " b =14 feet C = c = C = " c = 5 yrds A = 27 =12 B " = b =17 C =13.03 c = 5.96 A = " = A = 58 = 25 g. B = " b = 50 h. B = b = 30 C = " c = 38 A = 9 36 " =14.05 B " = " b = 50 Not Possible C = " c = Solve the following tringles using the Lw of Cosines (nd then the lw of Sines). A = = 22 A = = 99. B = 42 b =14.73 b. B =15.82 b = 41 C = c =16 C =123 c = A = = 28 A = =1 yrd c. B = b = 42 d. B = b = 2 feet C = c = 55 C =14.09 c =14 inches 6. Oblique Tringles Stu Schwrtz

16 A = = 4500 mm A = = 7 e. B = b = 5000 mm f. B = b = 5 C = c = 4000 mm c = 20 Impossible 5. Find the res of the following tringles. A = = 22 A = = 99. B = 42 b = b. B = b = 41 c =16 C =123 c = Are = Are = A = = 28 A = =1 yrd c. B = b = 42 d. B = b = 2 feet c = 55 c =14 inches Are = Are = in 2 A = = 4500 mm A = = 7 e. B = b = 5000 mm f. B = b = 5 c = 4000 mm c = 20 Are = 8,549, Are = Impossible A =151 =125 A = 58 = 25 g. B =17.84 b = 79 h. B = b = 30 C =11.16 Are = Are = Impossible 6. Oblique Tringles Stu Schwrtz

17 6. Two rods intersect t n ngle of 52.7 with field in between. A girl is wlking on one of the rods 1.5 miles from their intersection. Her house lies 0.85 miles from the intersection long the other rod. If she cuts cross the field to her house, how much wlking milege will she sve? d 2 = " 2 (.85)( 1.5)cos52.7 d =1.19 Distnce sved = "1.19 =1.16 miles 7. Kurt wnts to sil his bot from mrin to n islnd 15 miles est of the mrin. Along the course, there re severl smll islnds they must void. He sils first on heding of 70 nd then on heding of 120 (remember tht hedings ngle mesures rotted clockwise from the north). Wht is the totl distnce he trvels before reching the islnd? A = 30 = B = 20 b = C =130 c =15 = 15sin30 sin130 = 9.79 b = 15sin20 sin130 = 6.70 Totl distnce : miles 8. A blloon is sighted from two points on level ground. From point A, the ngle of elevtion is 18 nd from point B the ngle of elevtion is 12. A nd B re 8.4 miles prt. Find the height of the blloon if ) A nd B re on opposite sides of the blloon nd b) A nd B re on the sme sides of the blloon. = 8.4sin18 sin150 = 5.19 h = 5.19sin12 = 1.08 miles = 8.4sin12 =16.71 sin6 h = 16.71sin12 = 3.47 miles 9. A golfer tkes two putts to get the bll into the hole. The first putt rolls the bll 10.3 feet in the northwest direction nd the second putt sends the bll due north 3.8 feet into the hole. How fr ws the bll originlly from the hole? d 2 = " 2( 3.8) ( 10.3)cos135 =13.26 ft 6. Oblique Tringles Stu Schwrtz

18 10. Two mrkers A nd B re on the sme side of cnyon rim nd re 72 feet prt. A third mrker is locted cross the rim t point C. A surveyor determines tht "BAC = # nd "ABC = #. Find the distnce between A nd C. AC = 72sin51 38 " = ft sin " 11. A blimp is sighted simultneously by two observers: A t the top of 650-foot tower nd B t the bse of the tower. Find the distnce of the blimp from observer A if the ngle of elevtion s viewed by A is 31 nd the ngle of elevtion s viewed by B is 58. Find the height of the blimp lso. = 650sin121 = sin27 h = sin58 = ft 12. As I complete the Boston Mrthon, I view the top of the Prudentil Center building t n ngle of elevtion of 7 29 ". I run one mile closer nd now view the ngle of elevtion s ". How tll is the Prudentil building nd how fr wy from it m I now? = 5280sin7 29 " = sin3 5 9 " h = sin11 28 ". =1, ft x = sin11 28 " = ft =1.84 miles 13. A tringulr piece of lnd in prk is to be mde into flower-bed. Stkes hve been driven into the ground t the vertices of the tringle which we will cll B, E, nd D. The grdener cn only locte the two stkes t B nd E. BE mesures 6.2 meters, nd the grdener reclls tht the ngle t B is 60 nd the side opposite the 60 ngle is to be 5.5 meters in length. Bsed on this informtion how fr from B should the grdener serch for the missing stke? B = 60 b = 5.5 E = e = 4.29 ft D = d = 6.2 B = 60 b = 5.5 or E =17.49 e =1.91 ft D = d = Oblique Tringles Stu Schwrtz