Trig Part 3 Final Exam Review Key

Trig Part 3 sinA = si B sl C � = � Law of Sines: Laws, Inverses, Solving, and Areas of Triangles Law of Cosines: a2 = b 2 + c2 - (Zbc cos A) Det,rmlne wl1etl,er to use the law ofSina of the Cosines to solv, the 1tlv,n trlan1tla, Circle one. You do NOT need solve/ lo I. 2, A @ A � 8 C 39 Cosines Es) Cosines ,. �C US< tli, law ofSines an,Vor tl1t law of Cosines to solv, tach trlanglt, Round all decimals to lht ntaral lotr, 3. 4. 2J 1i�·i1���0 :46 A B __ 7 2' C 40" Sil.l - --'o s,�&f � SIV'l Si-1.1 ° h1�'o2 ID �; I o 5,i-1 :i4. 1° : s lVIA I•,: lt't'l LA= �,A· $lnit• 'o, i.i 0 LB= S4.I ° 5, LA= 32.2°, b = 21.3, a = 34.5 b= ��.'5 0 11.;�,11�t�•: 34.S"J1'(16 s,n �,:. i1:,$1Y13l.i• ,?,'/.? �= \"i.i - 0 c= So.Ii s1:i.=t ° zc;"- SlYI ll.:t . SI;.. c·-So.t, L8 = \ G\ ''J." LC= \'l,..'6,1,, 0 si"';i.1� ��l s,11�1..1• l":l-$1vitaSj' ,, s1.i C 21, 0 C = Llo,"t c- LC= Lfo,=t 0 6. LA= 127', b = 32, c = 25 o.,.: ;i."+ L.( �:z.)tz.i;)eos12.:r• 0.; i::i1. I 0 ;q1,.,,·11 1�.1}: csiil3 1• -C� '>'f,5_Sl� 113-�• - - s ot= u.11.C\011 �.'$ 51-ri ei; .�" -s ,� C 1:; 2,(, 0 1":h,;.. ,as ,'t -.. ZCPsil'lC -54 �:�B ::i-0 4 t"-lti.c,• ll,'.> --�- -54,, -:i-1� cos!? LA= S3,,; LB= 1io • ,1..i1q.�•" M,,t0 �,V"le SiYI --•- 4, A .. S 3.tl C 26 5 .. i u. 0 (p�•l.\�IH�- �1\\co-s6 10 t" s \Y\ ia2 ; I OS IYIA ° 21;:. 2.1"�11-� -,2.(1.1){.l-:t)c.ost:'> bs1Y1'bi" 1os,11si1.1• ° 0 o A S1Y161°.., S1-n A. .., I0 6 1?>0 • 0!:>,:t • 'io,"t 0 8 • l� -ll't 0 ?.. 3 :. 0 c. 4- i, ... 30 ° St.I � �s 2Ssin\i:t0:: ?1,1 �,� C - Slv-.C-- l5Sl� 11�• 1:)1.1 C-:-i;• LB= -'3Q LC= --').. ?: o 0 a= c; l,\ S.of�l!'•lol(owlng,gtlaphat.•If• s'8 s 3�0;•''!!!fi"!!!!lf/,f"!��°"llf1ln4,lf8/v!'I "!', 12. 2 tan 9 sin 9 -2 sin9=0 (S) 2s,.;..e(t4t19-i)-=-o o 13. 2cos2 9 -v'3 cos9 =0 cose(Z.cose -�),=u + + + 2s,viG-:o �&-,�o ) o�,���'4 45 ,'l-l�•J IS. 2cos9 +I=0 (2) + Z.cose�:o (90 , 110• ° ( 2..tii.,.,e-�)( to."e-1 ')=o 1120 , 2.'lo•j 0 2.tntle-3 :.0 = ¥z. + -bM0 � ·I =-0 ;;•J t�r.-�o) i% -ttu,.e :: I � ( 12-)(2-4')s, YI ten· o 2-z [P1ru..; l?ct \ � \ +A::. 1io· -102'. • n•= 51' --lo 8·5 St�A _ S1V' \Ot Slt"OI" � --':,.- s,.:-e::z. ,.g. -t 17. 2cos29=cos9 (4) 2GC5el ·C.058 :: 0 cose(.2c.45e -1)-:..o eose = o �se =i +. �o• ,�0° /9c!,2.�• i,o'l:.�,��·� ��f,y,9Jli��-1,ftnd ickifflcciil!fnt'ornlallc,ni��"� '· a =32, b =24, c=36 � c %2.-. ;2,2.-t 2.1\)..-2.(�2)(21-1;co.s \ 2,(\t,:: 102-'1 +- ?1-(, - IS¾ �sC, Cos.C-= � -\'S?lo ° /) L:. bto.-:r "3_ ( �2 "3,1V15"1 ,ktJJ.:. s,,.:..e: -1 ��·1 2 b ... 8.Ss,,;,10 ° At�tt�·. o - 30Ll -=- -1i; ?LI c.os C. 511"1\02.· i.S s ,;.?&-s ,;,e •2."' ( s,,:.,e-2.')(s,..-..e ... i') -=-o + + l4s , ,rj •(i• RNI ""-• l101!2'.:"'�"."'•, •'·sliee •---""' - d• � ;IDCS'Of�uJICIH! .aW\0 \C..I 19. B=102',C=27',a=v, 18. A=l05',b=12,c=24 "i 1,,c S1V1 A .�'!�1 14. sln29-sin9=2 (I) 3o",33o•/ 16. 2tan29-Stan9+3=0 (4) 2. CoS8:. -I Cose::-{ <•+'½ � c.ose�o 1M8 ° (4) n '5).1o::l)si" t1" '2.o.£,w,u+s1 -1 cos o '-\\ (·Sc, -=isiu)-= tS . la J(l.4) Stvi '¾ .lo� \wu.:. ; '¼. 4 w,;.�J Given u = (-4, 9), v = (-2, -5) and w = (7, 0). Find the following. 6. llvll 5. llull ✓c-"l)�+-C!' 7. llu+ vll ✓ I.( +Z.5 U+>J:::<-1., 14) ffl V3� +It, -./c-:1-')'·-t-c-sY· fii.+ �\ a � 8. llv+wll v-,.w, t0.} <20115 2v-t-?N � � Ii.+-(·'-)" ZO 0-(-IS)" Ii:; i,'·l; 11.) <-ti..5, 45'> 12. ll 3w- 4vll ?;w, < 211 o "l 4v :. <- �' • Zo-, (21 10 -,-<-s,-20) 2.1-(-'fr:- 2.'! 0-(•Zo); 20 '1' 1 '" 2.'! -t- 2.0 <2'1,20> 12.) 3S.Z. = Ji'i'i:TT t,.21 = Find the unit vector in the direction v of by dividing each component ofv by the magnitude ofv. -13. v- =-(8.-6) -J"ceo\��c-v'>" .. lto4.'"&. i 4- IO, 1;' - r. _?, io" � , .fioo .. 10 14. = (15, -8) c1r:.)2+<-e)a.= [22.s+�u,.,_JzeP..,, l't- V - ✓ }.i I� 1 -� I';- l,j --'- 13.) -> - � <. 5 > S IS -9 ,;: ) 14.)< � I I