Section 6.1 Law of Sines. Notes. Oblique Triangles  triangles that have no right angles. A c. A is acute. A is obtuse


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1 Setion 6.1 Lw of Sines Notes. Olique Tringles  tringles tht hve no right ngles h is ute h is otuse
2 Lw of Sines  If is tringle with sides,, nd, then sin = sin = sin or sin = sin = sin The miguous se (SS) given two sides nd n ngle opposite one of the sides h is ute is ute is ute sin = = sin = h < sin < h one solution no solution one solution is ute is otuse is otuse sin < < h < < two solutions < no solution > one solution
3 lthough some tehers wnt you to memorize ll these ses, I think it is esier to use ommon sense. Rememer tht in tringle the lrgest side must e opposite the lrgest ngle nd the smllest side must e opposite the smllest ngle. If everything doesn t mth up then you don t hve solution. re of Tringle  = 1 2 h onsider the right tringle = h Sine sin = nd sin = = h we n sustitute in the ove eqution nd get re = 1 2 h = 1 re = 1 2 sin This formul n e expnded to 2 sin re = 1 sin 2 These new formuls for re n e used when you hve two sides nd the inluded ngle of tringle.
4 Exmple Prolems. #2 pge 414 Use the Lw of Sines to solve the tringle. If two solutions exist, find oth In tringle the ngles dd up to = =180 = sin35 = sin55 18 sin55 Using the Lw of Sines = sin nd 18 sin35 = sin90 18 sin90 = sin #8 pge 414 Use the Lw of Sines to solve the tringle. If two solutions exist, find oth. = 60 = 9 =10
5 sin 10 = sin sin60 Using the Lw of Sines sin = 9 = sin 1 10 sin60 = =180 To find the missing ngle =180 = 45.8 sin45.8 = 9 sin60 To find the missing side = sin sin60 = ut wht you ould esily miss in this prolem is tht there is nother solution. We solved for ngle nd got 74.2 for solution, orret solution. Is it possile tht there is nother solution for ngle? YES! Use = 74.2 s your referene ngle in the seond qudrnt nd you get
6 So we hve seond solution with = =180 To find the missing ngle =180 =14.2 sin14.2 = 9 sin60 To find the missing side = sin sin60 = If you hek eh solution you ll see tht the lrgest side () is opposite the lrgest ngle () nd the smllest side () is opposite the smllest ngle (). I know this worries some of you euse you might not think to look for seond solution. DON T WORRY, when you lern the Lw of osines you will hve method tht will lwys tell you how mny solutions you hve. #20 pge 414 Find the re of the tringle. =130 = 92 = 30 Using the formul re = 1 2 sin we get re = sin130 = #30 pge 415 pilot hs just strted on the glide pth for lnding t n irport with runwy of length 9000 feet. The ngles of depression from the plne to the ends of the runwy re 17.5 nd () Drw digrm tht visully represents the prolem.
7 feet () Find the ir distne the plne must trvel until touhing down on the ner end of the runwy =1.3 nd lternte interior ngles gives us the following digrm x feet x sin17.5 = 9000 sin1.3 Using the Lw of Sines x = 9000 sin17.5 sin1.3 x 119, feet () Find the ground distne the plne must trvel efore touhing down x 18.8 ground distne = z x 119,
8 z sin71.2 = In the smller right tringle we hve z = x sin90 119, sin 71.2 sin90 z 112, feet (d) Find the ltitude of the plne when the pilot egins the desent. ltitude= y 71.2 x ground distne = z x 119, y sin18.8 = Using the Lw of Sines x sin90 119, sin18.8 y = sin90 y 38, feet #32 pge 415 ot is siling due est prllel to the shoreline t speed of 10 mi hr. t given time the ering to the lighthouse is S 70 E, nd 15 minutes lter the ering is S 63 E. The lighthouse is loted t the shoreline, find the distne from
9 the ot to the shoreline. ot d shoreline light house 15 minutes = 1 4 hour so during tht 15 minutes the ot trvels 10 mi hr 1 4 hr = 10 4 = 5 = 2.5 miles 2 Find ngles,, nd ot d shoreline light house = = 20, = =153 nd to omplete the tringle = = 7
10 ot 2.5 shoreline light house sin153 = 2.5 sin 7 Using the Lw of Sines = 2.5 sin153 sin miles ot d 20 shoreline light house sin20 = d Using the right tringle we get sin20 = d d = sin miles
11 Homework. pge 414/ 133 odd
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