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Vetors nd Olique Tringles SE 2: TWO SIDES ND THE NGLE OPPOSITE ONE OF THEM For tringle in whih we know two sides nd the ngle opposite one of the given sides, the solution will e either one tringle, or two tringles, or even possily no tringle. The following exmples illustrte how eh of these results is possile. 60.0 40.0 = 40.0 ' Side rehes t either of two points () Fig. 57 Fig. 58 60.0 () 60.0 ' ' () EXMPLE 4 se 2: Two sides nd ngle opposite Solve the tringle with the following given prts: = 60.0, = 40.0, nd =. First, mke good sle drwing (Fig. 57()) y drwing ngle nd mesuring off 60 for. This will more lerly show tht side = 40.0 will interset side t either position or. This mens there re two tringles tht stisfy the given vlues. Using the lw of sines, we solve the se for whih is n ute ngle: 40.0 ' 40.0 60.0 sin = 40.0 sin = sin -1 or Therefore, = 48.6 nd = 101.4. Using the lw of sines gin to find, we hve Thus, = 48.6, = 101.4, nd = 78.4. See Fig. 57(). The other solution is the se in whih, opposite side, is n otuse ngle. Therefore, = 131.4 = 18.6 Using the lw of sines to find, we hve = 180-1 + 48.6 2 = 101.4 sin 101.4 = 40.0 sin =180 - = 180-48.6 =180-1 + 131.4 2 sin 18.6 = 40.0 sin = 60.0 sin = 48.6 40.0 = 40.0 sin 18.6 sin = 25.5 40.0 sin 101.4 sin = 78.4 sin = 60.0 sin 40.0 This mens tht the seond solution is =131.4, =18.6, nd =25.5. See Fig. 57(). The omplete sequene for the lultor solution is shown in Fig. 58. The upper window shows the ompletion of the solution for,, nd. The lower window shows the solution for,, nd. n 311
Vetors nd Olique Tringles EXMPLE 5 se 2: Possile solutions In Exmple 4, if 7 60.0, only one solution would result. In this se, side would interept side t. It lso interepts the extension of side, ut this would require tht ngle not e inluded in the tringle (see Fig. 59). Thus, only one solution my result if 7. In Exmple 4, there would e no solution if side were not t lest 30.0. If this were the se, side would not e long enough to even touh side. It n e seen tht must t lest equl sin. If it is just equl to sin, there is one solution, right tringle. See Figure 60. Fig. 59 60.0 60.0 ' Side rehes ut too long for seond Fig. 60 60.0 30.0 Just touhes n miguous se Summrizing the results for se 2 s illustrted in Exmples 4 nd 5, we mke the following onlusions. Given sides nd nd ngle (ssuming here tht nd 1 6 90 2 re orresponding prts), we hve the following summry of solutions for se 2. Prtie Exerise 2. Determine whih of the four possile solution types ours if = 28, = 48, nd = 30. UTION SUMMRY OF SOLUTIONS: TWO SIDES ND THE NGLE OPPOSITE ONE OF THEM 1. No solution if 6 sin. See Fig. 61(). 2. right tringle solution if = sin. See Fig. 61(). 3. Two solutions if sin 6 6. See Fig. 61(). 4. One solution if 7. See Fig. 61(d). ' () () () ' (d) Fig. 61 NOTE NOTE Note tht in order to hve two solutions, we must know two sides nd the ngle opposite one of the sides, nd the shorter side must e opposite the known ngle. If there is no solution, the lultor will indite n error. If the solution is right tringle, the lultor will show n ngle of extly 90 (no extr deiml digits will e displyed). For the reson tht two solutions my result for se 2, it is lled the miguous se. However, it must lso e kept in mind tht there my only e one solution. reful hek of the given prts must e mde in order to determine whether there is one solution or two solutions. The following exmple illustrtes se 2 in n pplied prolem. 312
Vetors nd Olique Tringles n See the text introdution. Hvn Est 43.2 v pg 43.2 South Kingston v w 40.0 km/h v p 300 km/h Heding Fig. 62 Prtie Exerise 3. In Exmple 6, wht should e the heding if 300 km> h is hnged to 500 km> h? EXMPLE 6 se 2: pplition Kingston, Jmi, is 43.2 south of est of Hvn, u. Wht should e the heding of plne from Hvn to Kingston if the wind is from the west t 40.0 km>h nd the plne s speed with respet to the ir is 300 km>h? The heding should e set so tht the resultnt of the plne s veloity with respet to the ir v p nd the veloity of the wind v w will e in the diretion from Hvn to Kingston. This mens tht the resultnt veloity v pg of the plne with respet to the ground must e t n ngle of 43.2 south of est from Hvn. Using the given informtion, we drw the vetor tringle shown in Fig. 62. In the tringle, we know tht the ngle t Kingston is 43.2 y noting the lternte-interior ngles. y finding u, the required heding n e found. There n e only one solution, euse v p 7 v w. Using the lw of sines, we hve known side opposite required ngle 40.0 sin u = 300 sin 43.2 sin u = 40.0 sin 43.2, 300 known side opposite known ngle u = 5.2 Therefore, the heding should e 43.2 + 5.2 = 48.4 south of est. n If we try to use the lw of sines for se 3 or se 4, we find tht we do not hve enough informtion to omplete ny of the rtios. These ses n, however, e solved y the lw of osines s shown in the next setion. EXMPLE 7 ses 3&4 not solvle y lw of sines Given (se 3) two sides nd the inluded ngle of tringle = 2, = 3, = 45, nd (se 4) the three sides (se 4) = 5, = 6, = 7, we set up the rtios (se 3) 2 sin = 3 sin = sin 45, nd (se 4) 5 sin = 6 sin = 7 sin The solution nnot e found euse eh of the three possile equtions in either se 3 or se 4 ontins two unknowns. n EXERISES 5 In Exerises 1 nd 2, solve the resulting tringles if the given hnges re mde in the indited exmples of this setion. 1. In Exmple 2, solve the tringle if the vlue of is hnged to 82.94. 2. In Exmple 4, solve the tringle if the vlue of is hnged to 70.0. In Exerises 3 22, solve the tringles with the given prts. 3. = 45.7, = 65.0, = 49.0 4. = 3.07, = 26.0, = 120.0 5. = 4380, = 37.4, = 34.6 6. = 932, = 0.9, = 82.6 7. = 4.601, = 3.107, = 18.23 8. = 362.2, = 294.6, = 110.63 9. = 7751, = 3642, = 20.73 10. = 150.4, = 250.9, = 76.43 11. = 0.0742, = 51.0, = 3.4 12. = 729, = 121.0, = 44.2 13. = 63.8, = 58.4, = 22.2 14. = 0.130, = 55.2, = 117.5 15. = 4384, = 47.43, = 64.56 16. = 283.2, = 13.79, = 103.62 17. = 5.240, = 4.446, = 48.13 18. = 89.45, = 37.36, = 15.62 19. = 2880, = 3650, = 31.4 20. = 0.841, = 0.965, = 57.1 21. = 450, = 1260, = 64.8 22. = 20, = 10, = 30 313
Vetors nd Olique Tringles In Exerises 23 40, use the lw of sines to solve the given prolems. 23. smll islnd is pproximtely tringle in shpe. If the longest side of the islnd is 520 m, nd two of the ngles re 45 nd 55, wht is the length of the shortest side? 24. ot followed tringulr route going from dok, to dok, to dok, nd k to dok. The ngles turned were 135 t nd 125 t. If is 875 m from, how fr is it from to? 25. The loding rmp t delivery servie is 12.5 ft long nd mkes 18.0 ngle with the horizontl. If it is repled with rmp 22.5 ft long, wht ngle does the new rmp mke with the horizontl? 26. In n eril photo of tringulr field, the longest side is 86.0 m, the shortest side is 52.5 m, nd the lrgest ngle is 82.0. The sle is 1 m = 2 m. Find the tul length of the third side of the field. 27. The Pentgon (hedqurters of the U.S. Deprtment of Defense) is the lrgest offie uilding in the world. It is regulr pentgon (five sides), 921 ft on side. Find the gretest stright-line distne from one point on the outside of the uilding to nother outside point (the length of digonl). 28. Two ropes hold 175-l rte s shown in Fig. 63. Find the tensions T 1 nd T 2 in the ropes. (Hint: Move vetors so tht they re til to hed to form tringle. The vetor sum T 1 + T 2 must equl 175 l for equilirium.) Fig. 63 42.0 T 1 175 l rte 29. Find the tension T in the left guy wire tthed to the top of the tower shown in Fig. 64. (Hint: The horizontl omponents of the tensions must e equl nd opposite for equilirium. Thus, move the tension vetors til to hed to form tringle with vertil resultnt. This resultnt equls the upwrd fore t the top of the tower for equilirium. This lst fore is not shown nd does not hve to e lulted.) T 2 31. Find the distne etween Grvois ve. nd Jefferson ve. long rsenl St. in St. Louis, from Fig. 66. Fig. 66 Grvois ve. 1.12 mi 50.5 rsenl St. Stellite 0.88 mi Jefferson ve. 32. When n irplne is lnding t n 8250-ft runwy, the ngles of depression to the ends of the runwy re 10.0 nd 13.5. How fr is the plne from the ner end of the runwy? 33. Find the totl length of the pth of the lser em tht is shown in Fig. 67. Fig. 67 6.25 m 31.8 108.3 Refletors 34. In widening highwy, it is neessry for onstrution rew to ut into the nk long the highwy. The present ngle of elevtion of the stright slope of the nk is 23.0, nd the new ngle is to e 38.5, leving the top of the slope t its present position. If the slope of the present nk is 220 ft long, how fr horizontlly into the nk t its se must they dig? 35. ommunitions stellite is diretly ove the extension of line etween reeiving towers nd. It is determined from rdio signls tht the ngle of elevtion of the stellite from tower is 89.2, nd the ngle of elevtion from tower is 86.5. See Fig. 68. If nd re 1290 km prt, how fr is the stellite from? (Neglet the urvture of the erth.) Fig. 64 Fig. 65 Memphis T 330 mi 640 mi 136 tlnt 105.6 850 N 35.7 30. Find the distne from tlnt to Rleigh, North rolin, from Fig. 65. Rleigh Fig. 68 86.5 89.2 1290 km 36. n stronut on the moon drives lunr rover 16 km in the diretion 60.0 north of est from the se. () Through wht ngle must the rover then e turned so tht y driving 12 km frther the stronut n turn gin to return to se long north-south line? () How long is the lst leg of the trip? () n the stronut mke it k to se if the mximum rnge of the rover is 40 km? 314
Vetors nd Olique Tringles 37. ot owner wishes to ross river 2.60 km wide nd go diretly to point on the opposite side 1.75 km downstrem. The ot goes 8.00 km>h in still wter, nd the strem flows t 3.50 km>h. Wht should the ot s heding e? 38. motorist trveling long level highwy t 75 km>h diretly towrd mountin notes tht the ngle of elevtion of the mountin top hnges from out 20 to out 30 in 20-min period. How muh loser on diret line did the mountin top eome? 39. hillside is inlined t 23 with the horizontl. From given point on the slope, it hs een found tht vein of gold is 55 m diretly elow. t wht ngle elow the hillside slope from nother point downhill must stright 65-m shft e dug to reh the vein? 40. Point P on the mehnism shown in Fig. 69 is driven k nd forth horizontlly. If the minimum vlue of ngle u is 32.0, wht is the distne etween extreme positions of P? Wht is the mximum possile vlue of ngle u? Fig. 69 nswers to Prtie Exerises 36.0 m 24.5 m 1. = 6.34 2. Two solutions 3. 46.3 P 6 The Lw of osines Lw of osines se 3: Two Sides & Inluded ngle se 4: Three Sides Summry of Solving Olique Tringles () () h Fig. 70 x x h s noted in the lst setion, the lw of sines nnot e used for se 3 (two sides nd the inluded ngle) nd se 4 (three sides). In this setion, we develop the lw of osines, whih n e used for ses 3 nd 4. fter finding nother prt of the tringle using the lw of osines, we will often find it esier to omplete the solution using the lw of sines. onsider ny olique tringle for exmple, either tringle shown in Fig. 70. For eh tringle, h> = sin, or h = sin. lso, using the Pythgoren theorem, we otin 2 = h 2 + x 2 for eh tringle. Therefore (with (sin ) 2 = sin 2 ), 2 2 sin 2 x 2 (9) In Fig. 70(), note tht 1 - x2> = os, or - x = os. Solving for x, we hve x = - os. In Fig. 70(), + x = os, nd solving for x, we hve x = os -. Sustituting these reltions into Eq. (9), we otin 2 2 sin 2 ( os ) 2 nd 2 2 sin 2 ( os ) 2 respetively. When expnded, these oth give 2 2 sin 2 2 os 2 2 2 os 2 (sin 2 os 2 ) 2 2 os Relling the definitions of the trigonometri funtions, we know tht sin u = y>r nd os u = x>r. Thus, sin 2 u + os 2 u = 1y 2 + x 2 2>r 2. However, x 2 + y 2 = r 2, whih mens sin 2 U os 2 U 1 (12) This eqution is vlid for ny ngle u, sine we hve mde no ssumptions s to the properties of u. Thus, y sustituting Eq. (12) into Eq. (11), we rrive t the lw of osines: (10) (11) Lw of osines 2 = 2 + 2-2 os (13) Using the method ove, we my lso show tht 2 = 2 + 2-2 os 2 = 2 + 2-2 os 315