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HA1 VECTC)VA '&NG DUNG

0,Cho tam gidc AB C vu8ng

I ta i A va c6 BC = 2a,I CA = a. Hay tinh c6c 11I r6 bong gidc rau ddy :

6 1dp 9 ta di3 bigt dinh nghia ti so" ldung giBc clia g6c nhona = xOy nhd sau :La"ymat diem M b6t ki tr6n tia Oy sao cho M khbng trhng v6i0 r6i vE MP vubng g6c vdi t ia Ox. Ta c6 tam giAc OMP vu6ngtai P. Ta goi :

MP canh dbi-ins = --OM canh huydnOP canh k6-osa =-OM canh huykn /MP canh dbi ;tanga =-OP canh k& 0 POP canh kdcotanga =-MP canh d6 i' Hinh 2.7

CBc tl so"nhy kh6ng phu thuac vho tr i cSa diem M tr6n Oy(h.2.1). B&i v@y a c6 thz chon di6m M sao cho OM = 1 vhta c6 :

s i n a = MP , cosa = OPTa c6 the m&rang khdi ni$m ti sd Z~qng icic czia gdc nhon athhnh khBi ni$m gia trt Iuung giac czia mot g6c a vdi0 I 1180' bang dinh nghia sau dgy :

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I.Dinh nghia. Cho h@ og do Oxy vdi c6c didm A(-, ; O),I /i j B ( l ;0) v i C(0 ; ). Ta go i n8a d&ng trbn dllirng k inh AB vA! i d i qua didm C la naa d&ng trdn don vj (c6 b6n k inh bang 1).

Hinh 2.2Vdi m6i g6c a ( 0

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30 Chllmg II. T/CH HIJ~%JGC~IAA1VECTI3.

Ch o tam gi6c ABC.Chdng minh rang :sinA = sin(i3 + C)

J5 J5Vey sin 135"=- cos 135"= -- ,tan 135"= -1, cot 135"= -1.2 2

Hinh 2.3

2. Gia tr [ Itfdng giac c6a hai goc bu nhau

ia) Dinh / j . Hai g6c bC nhau c6 sin b%ngnhau vA c6 c8sindbi nhau.s i n a = sin(180- a )cosa = -cos(l 80" - a ).C H ~ N G INH

_4LiXy di6m M tr6n n3a d&ng trbn d m v j sao cho MOx = a r8i-ii'y di6m M' d6i xxlihg v6i M qua truc Oy th i xOM' c6 so" dobiing 180' - a (h.2.4). R6 r b g a hai 6ie"m M vh M' c6 tung dob h g nhau cbn hohnh do clia chung thi 66i nhau. Do 66

sin a = sin(180- a )cosa = -cos(180 - a )

Hinh 2 .4

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51. GIA TRI LUONG GIAC CUA MOT GOC 31

I ) HS.qud. Hai g6c bO nhau c6 tang dbi nhau v i chtang dbil a / n hau.tana = -tan(180 - a7 cota = -cot(180 - a).

V i du. Tim cAc gid t r i ldung gidc clja g6c 150".

1sin 150"= sin 30" = - f i; cos 150" = -COS 30" = -- .2 2 'J3tan 150" = -tan 30" = --3 ; cot 150" = -cot 30" = -&3. Gio tr j lwng giac c6a mat sd goc doc biet

Yo 1Chri J?. Ta c6 tan 90' = - - 18 khang xac dinh vAxo O

1X0 - - cGng kh8ng xdc dinh n8n trong bHng lugngot o0 = -Yo O

0

gidc ciia c6c g6c dac biet ta dcng ki hi& " ( I " de" chi giA trikh8ng xdc dinh 66.

30"

1-245-1x43

sin a

cos a

tan a

cot a

Mat so" g6c dac bi$t khdc nhu cdc g6c 120, 135"' 150" d&u c6cAc gid t~ 1-g giAc dwc suy ra til cac giA tA 1Uqng gi6c ciiaCAC g6c d$c biet dii cho.

O

1

O

II

45"

-2J5-1

1

60"

-521-2

f i1-J3

90'

1

0

II

O

180"

0

-1

0

II

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32 Ch~ung1. T/CH ~6H ~ G~ AA1VECTU...

Ta ludn lubn c6 :sinus1vi cos a 5 1 .

J3V i du. sin 120" = sin(180- 120") = sin 60" =-4. Cac he t h k giila cac gia tr i lwng giac cca mot goc

s Dinh 11'. Vdi moi g6c a ta d8u c6 :n0 iI sinaa) N ~ Uosa * 0 thi tana =-cosacosa .b) N& sina # 0 thi cota =-sinac ) sin2a + cor2a = I.CHUNGMINH

a) va b) : Til djnh nghia ta c6 :YO sin at a n a = - vdi xo # 0) hay tan a =-o cosa0 co s acota =- vdiyo + 0)hayco ta = -.Yo sin a

C) Tren n3a d&ng trbn ddn vi ta l6y di6m M sao choMOx = a . Goi MI va M2 la hinh chi& cria M lhn l ~ d trkn Oxva Oy (h.2.5).- -a c6 sina = OM2 va cosa = OM1.

Hinh 2.5

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51. GIA TR! LLQNG GIAC CUA MOT GOC 3 3

2. Cuc he thdc khcicI5 Djnh 1;. Nku cosa+O th i 1 + tan2a =T.I i

i ! cos ai ;ii 1I N 6 u s i n a t O th i 1 + c 0 t 2 a = ~ .sin aC H ~ N G INH

2 2 22 sin a cos a+sin a-a c 6 1 + t a n a = 1 + -- - 1--2cos a 2cos a 2cos a2 2 2cos a s i n a + c o s a-+ c o t 2 a = l +T- - 1--

sin a 2sin a 2sin a3. Ap dunga) Cho bigt tanx + cotx = 2, h8y tinh sinx.cosx.

2 2sin x cosx sin x + cos xtanx + cotx =- -- - - 1cosx sl nx si nx cos x sin x cos xNe"utanx + cotx = 2 thi 1 = 2.sin x cos x

1Ta suy ra sinx cosx = -2

2= - 2 -sin a2D o 6 6 A =7- 1 12cot a=-+-- 2 2 2cot2asin a sin a sin a

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5. Cac dong bai tep cd b i nDung 1. De" tinh cAc gia tn bung giac clia mat g6c khi bi6tmat gia t n lIfdng giAc c3a g6c 66.Phuungphdp :- Dga vAo dinh nghIa, chn tim tung dij yo va hohnh do xo c3a-i&mM tr6n nila dlllrng trbn d m v j vdi g6c xOM = a vh t 3 66ta c6 :

s ina =yo ; cosa = xo; tan a = - Xo; cota = -.Xo Yo

- Dga vAo tinh chgt : hai g6c bh nhau c6 sin bhng nhau va c6casin, tang, c6tang d6i nhau.- Trong mot so" t m h g hqp c6 the" dga vho ti so" do d8i cAc c g hc h mat tam gihc vudng, cu thd 1A do"iv6i g6c nhon a ta c6 :

d6is ina =-uyQnd6i

t a n a =- h.2.6)k& Hinh 2.6D6i vdi hai g6c phu nhau a vh 90' - a ta c6 :sin(90- a ) = cos acos(90- a ) = sin a

vh ne"u 0 < a I 90 thi tan (90'- a ) = cotane"u 0' I a < 90thi cot (90'- a ) = t a n a .- Sir dung he that cos2a + sin2a = 1 d6 tinh tofin.

1Vi du 1. Cho cosa = -- Hay tinh c6c gi6 tr i l l ldng gi5c cbn3la i clia g6c a v21 so shnh c6c gi6 tri n i y vdi so" 0.

2 1 8Ta c6 sin2a = 1- cos a = 1- - = - Vi s in a > 0 nen ta lgy9 92& sin as in a =- ti suy ra tan a =- :(-i) - 2 J i3 cosa 31 -ots = -- 1 .J2--=--t a n a 2 f i 4 '

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Vi a 18 g6c tir n6n cAc gi8 tfi luang giac cosa , tan a , cotad&u18 c8c s6 Zim, chi risng sina lh s6 dumg.Vi du 2. Cho bigt tana = -2 . HZy tinh c3c gi3 tr i l~w ng i6ccbn lai cGa g6c a .

Ta c6 t a n a = -2 < 0 nen a lh g6c tir, nghia 18 90" c a c 180'.Do d6 cosa < 0. (1)

1 ITheo he thclc 1 + tan2a =-ta c6 cos2a =cos a 1+ tan2a2 1 1hay cos a =- -1+4 5 ' 1Tir (1) v h (2) a suy ra cos a = --'

sin aTheo he thirc tan a =- a c6 sin a = cosa .tanacosa2 2 45hay sin a = (-5)(-2) =- -5 5

1 1cota =- &nta c6 cota = --tan a 2 'V i dl! 3. 6i6t tana = f i .3sina - cosaTinh bigu thQc A = sina + cosaCcich I. Vi t a na = fi > 0 n8n a lh g6c nhon, do 66cosa > 0.

1Tir he th&7 1 + ta n2 a ta suy ra :cos asin at a n a =- 4 3 4 6n&ns i na = cosa . tana = -.& =-.cos a 3 36 f iThay sina =- h cosa =- ho bigu thclc A ta tinh d ~ d c3 3

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C h h 2. Chia ti3 vA mau cba A cho cosa # 0 ta c6 :

Vi du 4. Tinh :a) sin23"+ sin21So + sin2750+ sin2870b) cos21 + c0s*78~ cos21 + ~ o s ' 8 9 ~c) Cho bigt sin 15" = J 6 - 2 . H5y t inh cdc gid t r i4

Imng gi4c cbn la i clja g6c 15".GIAI

a) Chli 9 rhng v i sin 3" = cos(90- 3") = cos87"vA sin 15"= cos(90- 15")= cos75"nCn sin23"+ sin215"+ sin275"+ sin287"=

(cos287"+ sin287")+ (cos275"+ sin215")= 1 + 1 = 2.b) Vi cosl2"= sin(90- 12') = sin78"vA coslo= sin(90- 1")= sin89"nen cos212"+ cos278"+ cos210+ cos289"=

(sin2780+ cos278")+ (sin2 89"+ cos289")= 1 + 1 = 2.

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Dung 2. Chdng mi n h cac he thdc v6 gia tri l ~ u n g iac cua matgoc a .- Dqa vho dinh nghia gi6 tri ldung gi5c ciia mot g6c a vdi0 I 1 180'.- Dqa vao dinh li : "T6ng ba g6c c6a mot tam giac lu8n lu8nb%ng 180'".- Sd dung cAc he th~Ic u b6n dude suy ra til d inh nghia n h ~

sin a cos at a n a =- di cosa # 0, cot a =- di sina ;t 0,cos a sina2 2sin a + cos a = 1.

- Ap dung he thl'rc :2 1+ N b os a + 0 thi 1 +tan a =7

cos a1+ Ngus ina + 0 thi 1+ cot2a =- -sin a

- Thqc hien c6c phbp bi6n d6i t ~ n g ~ n g ,lra he th13c cAnchQng minh vd mat he thirc duuc thaa nh an la dling.V i du. ChQng minh r i n g :

a) sin4x + cos4x = 1 - 2sin2x cos2x ;b) sin6x + cos6x = 1 - 3sin2xcos2x.

I = 1 - 2sin2x cos2x (dpcm (I' )2= sin4x+ cos4x- sin2x cos x

= 1 - 2sin2x cos2x- sin2x cos2x= 1- 3sin2x cos2x (dpcm)

V i du 2. ChQng min h r%ng1-s ina cosaa) -C O S ~ l + s i n a

b) t a n a - s i n a - 13sin a cosa ( l+ cosa) '

(*) dpcm : di6u phdi chOng rninh.

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38 C h m g I. T~CHa U ~ GA HA1VECTU.. .

21- sina (1- sin a)( l+ sin a ) 1- sin aa> - -cosa cosa(1+ sin a) cosa(1+ sin a )0

- cosya - cosa- (dpcm).cosa ( l+ sin a ) 1+ sin as i n a .-- sin ab) tan a - sin a - cos a--. n

sin a-sin5a sin' a sin5a- 1- cosa- - 1 m).cb s a sin2 cos a ( l+ COS a,

V i du 3. Chilng minh r ing :

2 2s i n a c o s a- =sinacosa .b ) ' - l + c o t u ~ + t a n a

a, ~ + c o s ~ [ , - ( l - c ; a ) ~ ] - l + c o s a ( 1-2cosa+cos asin a sin a s i n a I,'- sin2 a J- l + c o s a ( 1 2cota 2 )+--s i n a ( I -= ssrna co t

- 1+ cosa- 2 2cota-2 cot a +sin a sin a- 1+ cosa .2cot a(- 1 --) osasin a s i n a s i n a- 1+ cosa 1- cosa (1+ cosa)(l- cosa ).2cot a. - 2 .2 cotasin a sin a sin a

2sin a 2cos a 3sin a 3- - cos ab) l - = 1-l + c o t a l + t a n a s i na + co s a s i na + co s a3 3sin a + cos a= 1- sina + cosa

2 2= 1- (sin a+ cos a - s inaco sa )= sinacosa (dpcm).

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51. GIA TRI L m G GIAC CUA M?T GOC 39

V i du 4. Chl?ng m inh chc b i@uhirc sau khbng p hu thuoc vho x.a) A = (sinx + cosx)' + (sinx - C O S X ) ~b) B = sin6x + cos6x + 3sin2x cos2x .

a) A = (sinx + c o ~ x ) ~(sinx - cosx12= (sin2x+ cos2x) + 2sinx cosx + (sin2x+ cos2x)

- 2sinx cosx = 2 .2b) B = sin6x + cos6x + 3sin2x cos x =

V i dl! 5 . Cho tam gidc ABC. ChQng m in h rhng :a) sinA = sin(B + C) ;b) cosA = -cos(B + C) ;

A + B Cc) sin- = cos- .2 2

A + B CC) sin-= cos2 2V i du 6. Tam gi5c ABC c6 A6 = AC = a vA c6 g6c nhon

= 2 a . Go i A H v3 BK l A cAc d&ng cao ve ta cdc d inhA vA B.a) Tinh do d;ii cdc dddng cao A H vA BK the0 a vA a .b)Chll lng m in h r3ng sin2 a = 2sin a cosa. A

GIAI- -) Ta c6 BAH=HAC = a (h.2.7)AH-- - cosa a AH = a cosaAB a

Hin h 2.7

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40 Chlrong II. T/CH vC) HU~NGGA HA1MCTU ..b ) Goi SAHCa di$n tich tam ABC, ta c6 :

2SAUC AH.BC = BK.AC t rong 66 BC = 2BH = 2.a sina.

Tti (1)va (2) t a suy r a sin2 a = 2cosa sin a (dpc

1 . H i y ph6t bi6u dinh nghTa gi6 tri ldung gidc clja 1n6tg6c a vdi0" 5 a 5 180".a) GiA si r cho g6c a = 120". Hay cho bigt c6c gi6 tri lclung gidc :sin 120", cos 120", tan 12OU, cot 120'.b) Vdi nhCfng gid tri n i o cila g6c a ta c6 :

- sina v i cosa cirng da"u ?- sina va cos a khdc ddu ?

2. Goi a l a so" do clja g6c MOx tr@nnda dclbng trbn dun vi. Hayx6c dinh VI tri ciia digm M tr@nnrlla dddng trbn dun vi d6 saocho :

1a) cosa = - . H i y tinh so" do cGa g6c a d6.21 -) sina = -. Khi d6 g6c MOx l a g6c nhon hay g6c tir !2

1 -) cosa = --. Khi d6 hi y ti nh s6do cira g6c MOx .2d) sina dat gi6 trj Idn nha"t.

3. Hay so sdnh c6c cap gi6 tri lcr~lng i6c sau ddy :a) cos 45" v i cos 60" ;b) sin 20" v i sin 60" ;C) cos 120" va cos 150' ;d) sin 135" v i sin 170".

4. Dua vao dinh nghia gi6 trj b ung gidc clja mat g6c a(0 5 a 5 180"), h i y tinh gi6 tri c6c bi6u thOc sau (vdi a, b, c lanhang so" cho trudc).a) A = a.sin 0" + b.cos 0" + c.cos 90"b) B = a.cos 90" + b.sin 90" + c.sin 180"c) C = a2.sin 90" + b2.cos90" + c2.cos 180"

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5 1. GIA TRI L W G GIAC C l j A MOT GdC

5. Kh6ng dirng m6y tinh b6 tcii, h i y tinh gi6 t r ~ 6c bi6u thrSlc sau :a) M = 3 - sin2 90" + 2cos260" - 3 tan245" ;b) N = sinx + cosx khi x bang 0" ; 5" ;60";C ) P = 2si1ix+ cos2x khi x bang 60" ;45" ;30" ;C) () = sili'x + COS'X khi x bang 135" ; 180" ; 160" .

6. Chifng minh c i c hhng d ing thOc :a) (sinx + cosx)' = 1 + 2sinx cosx ;b) (sinx - cosx12= 1 - 2sinx cosx ;C) sin4x + cosJx = 1 - 2sinzx cos'x ;d ) sinx cosx (1 + tanx) ( 1 + cotx) = 1 + 2sinx cosx.

7. Cho tam gi6c ABC vuBng tai C v i c6 = 34') canh AC = 10,5.H ay tinh cac canh AB, BC v i gbc B clja tarn giac.

8. Cho tam giic ABC vu611g tai (CA = 31,G. H i y tinh canh AB va c,

I va cc6c goc /

j canh BC = 12,3,i, c ~ i aam gi6c.

9. Chirng minh ranga) Vdi g6c u (0' 5 a 5 90') ta c6 :

sin 190"- a ) = cos acos (90" - a ) = sina ;

b) Vdi g6c a (0" c u I 0") ta c6 :tan (9U" - a )= cot a ;

C ) Vdi g6c a (0' I c l < 90') ta c6 :cot (90" - a 1 = tan a .

10. Chdng minh ri ng 1 2 22 = ta n a + c o t a + 2 .sin acos a

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42 Chumg II.~ C Ho HUI%JG C ~ AA1VECTCJ.. .

1 . Goc gi3a hai vectd8 ,Ch o t a m g i j c AUC vuGngc j n tai A ci, I la trungc1ic"mc d a c a n h BC .

Trong t ru i l ng ha p n i orich v6 h u d n p ;. 6 cOgia tr l cfuong, co gia t r ijrn, bang O !

r a) D j n / l , lg / l ja . Cho hai vectd a v i d6u kh lc vectd 6 . T il-at digm 0 a"t ki ta vP = i v i = g . G 6c AOB vd i sbdo t i l 0' dgn 180' d uuc goi la g6c giGa ha; ve c to a V A g .1 Ta k i hi& g6c g i h hai vecta a v i i; 18 ( a , 6 ).

bl C/,.ri3 . Cdch xdc dinh g6c giaa hai vecta a vB 6 theo d/nhnghia n6u tr6n khBng phu thuoc vAo viec chon digm 0. Hunnaatacbnc6( ; , G ) = ( g ,i).N6'u ( a , g ) = 90" hi ta n6i rhng hai vecta a vh $ v u d n g g d cvd.i nhau vh ki hieu a I .C ) V i d u . Cho tam giAc ABC vu6ng tai A v i c6 gbc B = 40".Khi 66 ta c6 :- - --(BA,BC) = 40'; (AB,BC) = 140'

Hinh 2.8

2. Djnh nghia tich v o hddng c6a hai vectda ) Di nh n g h i < ~ .ho ha i vecta a va d&u kh lc vectu 6k v6 hu'dng cda ha i vecto a vd la met sb, k i hi& l a

I d d ~ cdc d ~ n h di cirng thirc sau :

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i,

Vdi hai sd thuc J , b bStki , ta c6 (a.blL= a'.bL.-.-vby vdi hai vecta a,bb i t ki , t l jn g thUc :

Tr~rirng up n6u c6 mot trong hai vectu a vh bhng vectu 6-.-ta quy udc a . b = 0.b ) CIlri 2. Trudc diiy ta d5 bi6t t6ng cua hai vectu la matvectd, tich ccia mot s6 vdi mot vectu 18 mot vectu. d di y ta c6tich v8 hudng cua hai vectd la nzQt sd chd khdng phai la matvectu. Vi v$y n6n ngu&i ta goi tich 6 6 la tich v6 h ~ d n g ~ia aivectd.c ) V i du. Cho tam gi6c d6u ABC c6 canh a va trong tdm G.Tinh c i c tich v6 hlldng sa u ddy :

Th eo djnh nghia ta co (h.2.9):-- 1 2AB.AC= a.a. cos 60' = -a2-7 1 2AC.CB = a.a. cos 120' = --a A2 AI c6 dung kli61ig ?

Ta cluy ~ldc crn vcctuvubng goc vd i rnoi vecw

v i u . a = o .

-- 43GA.BA =- a".a. cos 30 =-2 Hinh 2.9d j Di6u kien vudng g6c ciia hai vectuTit djnh nghia tich v6 h ~ d n g ua hai vectu vB ta nhanthgy r in g :- - - -- N ~ Ua , b ) = 90" thi a. b = li t .I61 .cos 90' = o ;- - -. - -- N gu gc la in bu ta c6 a . b = O t h i d o a * O , b * 6 t a s u y r aDo 66 ta c6 dinh li :"Ridrr h i e n cdn u& dii d6' hai vecta a7 t 6 vci 6 + 6 ou6ng gcicu u i nlzau la tich v6 hudng ciia chiing h&ng 0".

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e ) Nh&n z4t. Trong v(t.li ta bi6t rhng nth c6 met l w f? tAcdong lGn mot v$t lhm cho v@t 66 di chuy6n tii 0 6e"n 0' th ic6ng A cba lvc F duuc tinh the0 c6ng that :

4 -A = IF^ .l0o11.oscp . (h.2.,10)trong 66 (f?1 lh culrng 60 c6a l~ tinh bhng Niu-ton (vigt tdt-la N), 100 1 1s do dM quang d u h g 00' tinh bhng m6t (vigt td t-a m), cp lh g6c gi3a hai vectu vh 00 ' tinh bhng 60, cbnc6ngA tinh bhng Jun (vigt tt$t18 J).

dTmng tofin hgc, giA t* cba bigu thm A = IF^.100 '1. cos cp kh6ng-k& d m v j do dmc gpi 18 tich va h u h g cQa hai vectcr @ "21 00 .f

0 0'Hinh 2.7

3. Binh phlldngv6 hllang

dai cria vecta dd"

8 - - 4 - 4 -

4. Cac tinh ch6-t c6a tich v6 hlrCjng

CH~ ] :

Ng&i ta dii c h h g minh duuc cAc tinh chgt sau dAy cba tich v6hudng :

Ngu a = b t ac 6 a . b = a . a = a . a cosoO= a .

4 - - .V6i ba vecta a , b, c bgt it vii moi so"k ta c6 :- . - + - ,a . b = b. a . (tinh chgt giao hofin)-. -. -t -.- - - .a .(b+ C) = a . b + a . c .(tinh chgt phan phG ddi vdi ph6p ceng vectu) ;

1-1 1-1 1-l2Vdi m ~ ioan thang ABta lu8n lubn c6 :

A B ~ z2 [q2s A B = @ .

Khi 66 tich v6 hudng a .a d@c ki hi& 18 a 2 .-2Sd a nhy dduc goi 16 binh p h m g u6 h ~ h gcLa vectu i .

Do 66 ta c6 djnh li :"Binh plzuung v6 h ~ d n gria mot tlecto b&ng binh p h ~ c r n gd o

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92. T~CHO H ~ GUA HA1VECTCY 4-5

Tif c8c tinh chgt cclja tich v6 hudng, ta c6 th6 su y ra cAc h$th3c sau diiy :-2 - - -2( i + G l 2 = a + 2 a . b + b (1)

- + 2 '2 + - + 2( a- b ) = a - 2 a . b + b (2)- - - - -2 -2 - 2 - 2(a + b) . ( a - b )= a - b = la1 -1bl (3)5 . Cac bai toan ap dungB i i to6n 1. Cho hai v e c t ~ vA 6 bgt kki, g ~ i ; la hinhchi&" cda vecta 6 tren d&ng thang chQa vectd a . ChQng

-,* + -minh rin g : a . b = a . b l

a) b)Hinh 2.7- -.T a vG ciic vectd OA = a vh = g . Goi B' 1h hinh chi& cclja Bt r e n dlrZIng th hn g OA. Ta c$n c h h g m inh :-- --OA.OB = OA.OB1 (h.2.11a) vA b)).-Ne"uAOB < 90' t a c6 :CI --.= 0A.OB.cos AOB = OA.OB .C4- Ne"u AOB > 90" th i : - --= OA.OB'.cos AOB = OA.OB .

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46 C h m g II. T~CH H W G A HA1VECTU...- -- N6u AOB = 90" th i OBI = 6 vB kh i d6-- --, A-OA.OB' = OA .O = 0 = OA.OB cos 90" = OA.OB .

BPi to6n 2. Cho doan thdng AB c6 do dAi 2a. Tim tap hdp-_.c6c di6m M sao cho MA.MB = k2 la mot s6 cho trddc.

Goi 0 a tru ng digm ciia doan th hn g AB (h.2.12) ta c6 :-- - _ . A -MA.MB = (M O +OA) (MO +OB)

Hinh 2.12

Viiy t e p hap cAc digm M lh d&ng trbn tarn 0 , bAn k inhR = m .B i i to in 3. Cho dddng trbn (0 ;R) vA dikm M c6 dinh. Matdddng th3ng A thay dbi lu8n lu6n di qua M cdt dlrirng trbn d6tai hai dikm A vA B. ChQng minh r ing :--M A . M B = M O ~ - R ~ = ~ ~ - R ~vdid=MO).

C H ~ N G INH -VE d ~ h ginh BC ciia d ~ & n gr b n ( 0 ; R). T a c6 MA 1B h inhchi& c6a MC t re n dudng th& ng MB. Theo cbng that h inhchie"u (BBi toftn 1) a c6 :

M / BHinh 2.73

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52. T~CH O H ~ GUA HA1 VECTIT 47- - - - - -A .MB =MC.MB = (MO + OC).(MO + OB)- -hri .i.. Gig tri khdng d6i clia tich v6 hlldng MA . MB =d2 - R~

n6i trong brii to6n 3 goi la phmng tich cua di6m M d6'i vd i

- Khi MT la tidp tuygn cuadie"m,ta c6 : trbn (0 ;R) vdi T la ti6p

Hinh 2.74

6. ~ i d uhilc toa do cGa tich v6 hdtrng-. -.Trong m8t phi ng toa do (0 ; i , ) cho hai vectd :

Khi do ta c6 t ich vi, h ~ d n g. .

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48 c h m g 1. WH~ f lUGNGC~IAAI VECT [I...

0,a) Cho a = (1 ; 2) ;6 = (-1 ; -3). Haytinh g6c ( a , 1b) Cho c = (2 ; 3) ;a = (3 ; 2). Ha y tfnh@ c ( ; , i ) .

{GI5fh o tam gidc ABC b i kA(1 ; I) , B(2 ; 3),C(5 ; -1 ). Hay tinh d8dAi cAc canh AB, BC,CA vA suy ra tam gi6cABC vu8ng tai A.

7. Ong dung c3a tkh vb hlrCInga) Do dai c6a vectd a = (al ; a2) d ~ u cinh theo cdng thtlc :

4b) G6c gigs hai vectu = (a l ; a2) vA b = ( b ~bz).Theo dinh nghia tich v8 h d n g ta c6 :

V i d". Cho = (- 2 ; I ) , = (3 ; 1 ) . Tinh g6c MON.- -- @.ON -6+1 &cosMON = cos(OM, ON) = - - --l o ~ ~ . l ~ l = & . f i -'- -.Vey (OM, ON) = 135'.c) Khozing c6ch giaa hai di6mCho ha i di6m A(xA; yA) ,B(xB; yB).Ta c6 = (xB - XA ; y~ - yA), do 66 t a t i n h d ~ u c hoHngc6ch giaa hai 6ie"mA va B 18 :

V i d l . Cho hai dism M(-2 ;8 v5 N(l ;0). Tinh doan MN.

-Ta c6 MN = (3 ; - f i vB khozing c6ch giaa hai di6m M, N 18 :M N = I E I = , / ~ = ~ .

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Cho tam giac ABC c6BC = a, CA = b, A6 = c.Chon he tryc Oxy sao chog6c 0 t r h g v6i C, di&m Bnam t ren phdn d~ a n g ~ i atruc Ox cbn di6m A c6tung dd duang. Chdngminh rang :

d ) Bbi todn. Cho tam giAc ABC bi6t A(2 ;3). B(8 ; 6& + 3) vAC(2 + 4 4 5 ; ). -Tinh do dcii cAc canh AB, AC v2 so"do clia g6c BAC .2) Tinh d ien t ich tam giAc ABC.

-- Z.A6 2 4 6 + 2 4 &v8 cos BAC = cos(AB, AC) = 12.8

Vi3y BAC = 30'.2) Goi BH 18 d&ng cao c3atam giAc ABC, (h.2.15). Ta c6

CIBH = AB sinBAC

Goi S 1A dien tich tam giAc ABC, ta c6 :

8. Cac dang bai tap cd b6nDqng 1. n n h tich v8 hadng cfia hai vectaPhrlmg phap.- DDng dinh nghia tich v8 hudng va cAc tinh chgt c3a tich vdh ~ d n g3a hai vecta.- Sir dung cAc hhng dhng thGc v6 tich v8 hu6ng.-.- --.- Dhng cdng thflc hinh chi& : AB.CD = A ' B .CD vdi A', B' l$nl ~ q t8 hinh chi& c3a A, B l$n giA cBa CD .

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50 C h m g II. ~ C Ho H ~ G~ > A A1 VECT(I...V i du 1. Tam gidc ABC c6 AC = 9cm, CB = 5cm, C = 90".

Tinh :- -a) AB.AC ;-4b) BA.BC .

a) Cach I . B-_. - -AB.AC = IABI . , I A c ~ cosA

AC= AB.AC.-B= A C ~ g2 = 8 1 (h.2.16)

C ACdch 2. Ta c6 C la hinh chi& cba B 1Cn giA cfia KC.Uo d6

---. -.-AB.AC = AC.AC = A C ~ 8 1 .-- -.-.b) BA.BC = BC.BC = B C ~i 3 8 hinh chi& cba BA tr6n--gih cba BC . Viiy BA.BC = 52 = 25.V i du 2. Tam giAc ABC c6 AB = 5cm, BC = 7cm, CA = 8cm.

--a) Tinh AB.AC tCf d6 tinh gid tri clja g6c A.-+b) Tinh CA.CB .X6t tam gihc ABC ta c6

-2 - - -2 -2 --BC = (AC - A B ) ~ AC + AB - 2AC.AB.

AB.AC 20 1 -Ta suy ra cosA = --- - - nCn A = 60".IEI.IzI - 5.8 2b) TUcfng tg nhu cAu a), ta c6 :-- 1 1

CA.CB = - C A ~ CB2 - AB2) = - 82 + 72 - s2) 44.2 2

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1 Ej2. T~CH O H ~ GUA HA1 VECTU 5 1

V i du 3. Cho tam gi6c ABC vu8ng 6 A, Cc6 g6c = 60' vA canh huydnBC = 6cm. Tinh tich v6 h ~ b n gfia bc6c c$p vecta sau d i y :--a) CA.CB;b) AB.BC.

h 6o0AA B

GIAI HinTam g i l c A BC 18 t a m g i l c v u 6 ng c 6 = 30" ne n

1AB = - BC = 3cm2AC = 4 = 436-9 = f i 3& (h.2.17)-- 43a) Ta c6 CA.CB = CA.CB.cos 30" = 3 4 3 . 6 . = 27.2

Vi dl? 4. Cho hinh vu8ng ABCD canh a vh c6 t l m 0.'Gpi Mla mot digm tujl 9 tren canh BC. HZy tinh cAc tich vBhddng :--) MA.AB ;b) E .G .

b) Goi H la t r u n g didm cGa canh AB.Ta c6 OH I B. Do 66 :

a) Di ing can g thi lc hhih c h i& t a c 6-- --M.4.A.B = BA.AB = a.a.cos 180"

Hinh 2.18

l

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52 Chlldng I I . flCH HU%G ~ AA1VECTC7..2

Dang 2. C h h g minh cAc dang that v6 vectd c6 lien quan d6ntich v6 hudngPhuung phdp .- Sfi dung tinh c hgt ph5n ph6i ccria tich v6 h ~ d n g 6i vdi ph6ptong cfic vec tu.- Dhng quy t$c ba di6m d6i vdi ph6p ceng hoec tril vectu, vidu n h ~6i vdi ba di6m A, B, C ba"t kl, a luBn c6 :

V i du 1.Cho tU giAc ABCD bgt kki. ChUng minh ring :

-.- --t -. -..- -.ADA.BC = DA.(DC-DB) = DA.DC - DA.DB (1)-.- --. - -- --DB.CA = DB.(DA - DC) = DB.DA - DB.DC (2)-- - - _. _.A -_.DC.AB = DC.(DB-DA) = DC.DB - DC.DA (3)V i dl! 2. G Q ~ Ih trung digm clja doan thiing AB v3 M I3 mot

di6m tujl v . ChUng minh r3ng :--.MA.MB = OM^ - O A ~ OM^ - O B ~ .Vi 0 1h trung die"rn cira doan AB n6n = -s-- - - - -a c6 : MA.MB = (OA - OM).(OB-OM)- - -- OB + OM).(OB- OM)Vi du 3. Cho hinh ch2 nhat ABCD vh M Ih mot di6m tuj/ v .

ChQng minh r3ng :a) MA^ + M C ~ M B ~ M D ~-- --b) MA.MC = MB.MD .

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GIAa) Goi 0 18 giao 6ie"m ciia hai d&ng chBo AC va BD cSa hinhch3 nhet. Ta c6 :- - -M C ~ (MO +O A ) ~ (MO +oc12-_. -2M02 + OA2+ OC2 + 2 MO.(OA +OC) .- -2 ~ 0 ~ + 0 ~ ~ + 0 C ~ , ( v iA + O C = G ) (1)M B ~ + M D ~ =M ~ + S I ) ~ + ( E G + O D ) ~- - -2M02 + O B ~ OD2 + 2 MO.(OB + OD) .- -.= 2 ~ 0 ~ + 0 ~ ~ + O ~ ~ , ( v iB + O D = ~ ) 2)Vi 0 la t6m c6a hinh ch3 nhet n6n ta c6O A = O B = O C = O D .So sdnh (1)vh (2) ta suy ra :

M A + M d = 2 M d- -B + M D = ~ = .- - - -.Do 66 (MA +M C ) ~ (MB +O D ) ~

I Hinh 2.79VS du 4. Cho tam gi6c ABC vdi ba trung tuye"n Ib AD, BE, CF.

ChQng minh rang :

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Vi AD, BE, CF 121 c6c d ~ a n grung tuygn n6n ta c6 :

- - - - -- BC.(AB+ AC)+ CA.(BA+ BC) + AB.(CA + =)I2Dang 3. Chimg minh sg vu6ng g6c cOa hai vectdP h m g phap .Sd dung tinh chgt c6a tich v6 hrldng : i d i . 6 = 0.V i du 1.Chang minh trong met tam gi i c ba d&ng cao d6ng quy.

Gi6 s 3 tam giac ABC ba"t kk c6 hai d&ng cao BB' vh CC' c i tnhau tai H. Ta c6n ch1311g minh AH vu6ng g6c vdi BC (h.2.20).

-- - -. -.-. --A.BC = HA.(HC -HB) = HA.HC-HA.HB (1)-- - - -. -- -.-HB.CA = HB.(HA - HC) = HB.HA -HB.HC (2)-- ---. - - . .-HC.AB = HC.(HB-HA) = HC.HB -HC.HA (3)--i HB I CA n6n HB.CA= 0 vh--C I AB n6n HC.AB= 0. Do d6 tir (4)--a suy ra HA.BC= 0 hay Hd I Bd ,nghia 18 ba d&ng cao ciia tam giac CpABC dbng quy tai H. 1

II

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Vi dl! 2. Tam gi6c ABC czn c6 AB = AC. Goi H I2 trung di6 mcira canh d i y BC vA D l a hinh chi& cl ja H tr6n canhAC. ChQng minh r i n g du'dng trung tuye"n AM clja tamgi6c AHD vu6ng g6c vdi du'ang than g BD.

--Ta c dn c h h g m in h AM.BD = 0 (h.2.21).

Hinh 2.21Vi M 1h trung digm c6a H D n 6n ta c6 :

A _ . -2 A M = A H + A DA _ . _ .vA B D = B H + H D-- - - - -o d6 2 AM.BD = (AH + AD).(BH + HD )-- - - -AH.HD + (AH + HD).HC- - - - ---= HD.(AH + HC ) (vi H C = BH,AH.HC= 0).

Vey AM Vtiang g6c vdi BD .V i dl! 3. TO gi6c ABCD c6 hai d&ng ch4o AC va B D c$t

nhau tai 0. Goi H, K I$ n lddt I2 truc t zm cl ja c6c tamgi6c ABO, CDO v2 goi I, J the0 thO tu la trung di6 m cljaA D VA BC. ChOng minh rang H K vu8ng g6c vd i I).

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56 Chlrmg 11. T~CH H U ~ J G L~AHAI VEC U...

Hinh 2.22--Ta c6n c h b g minh H K . I J = 0.-- .--- --a c6 2 HK.IJ = HK.(AC +DB) = HK.AC +HK.DB-- - - - - *Do 66 2 HK.IJ = AC.(BD + DB) = AC.0 = 0.

V@yHK vu8ng g6c v6i IJ.V i du 4. Cho hinh thang vu8ng ABCD c6 ddirng cao AD = h, hai

canh d5y AB = a v5 CD = b. Tim he thilc giOa a, b, h saocho dlrErng cheo AC vu8ng g6c vdi ddirng ch6o BD.

(H.2.23) --a c 6 : ACIE o A C . B D = O- - -C.(AD -AB)= 0-- --.cs C.AD - AC.AB = 0 (1)

Hinh 2.23-- 2Ta c6 AC.AD = AD = h2 (cBng thGc hinh chigu)--. --C.AB = DC.AB = ab (cdng thGc hinh chi&)-- --Do 66 I% o C.AD - AC.AB = 0

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Dung 4. BiGu thirc t o a di) clia tich va huung v9 cAc irng dung :tinh 66 dhi ciia met vecta, tinh k h o a n g cBch giaa hai die"m,tinh g6c giaa hai vecta.

- Cho vecta G = (al ; a2)vii = (bl ; b2).Khi 66 :-.- Cho vecta u = (u l ; u2).Khi 66

- Cho hai di8m A = (xA yA)va B = (xB ; yB).2 2- xA) + (yg - yA) .

+- G6c gi3a hai vectu = (al ; a2) va b = (bl ; b2) d ~ q cinhthe0 c6ng that :

V i du 1. Trong mgt ph%ngOxy cho ba di6m A(3 ;5), B(-5 ; ,C(0 ; 4).a) Tinh do dai cdc canh AB v3 AC cfia tam gi5c ABC.-) Tinh g6c BAC .-ay BAC = 45".V i du 2. Cho tam g i ic ABC bigt A(-3 ;61, B(l ; 2), C(6 ; ).a) Tim toa do trong tdm G c i a tam gidc ABC.b) Tim toa do truc t3m H clja tam gi%cABC.

G I ~a) Ta bigt trong t6m G c6a tam gihc ABC c6 toa dij 18 :

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b) Goi H(x ;y) 18 trgc t h am giac ABC, ta c6 :

Vey trgc tam H cda tam gihc ABC c6 toa de 1& (2 ; 1).V i du 3. Cho hinh vu8ng ABCD c6 canh bsng a. Goi E I5

trung di6m c i a canh AB, F la di6m trEn canh AD sao1-cho AF =-AD. XAc dinh v i tri clja di6m M tren d&ng3

th ing BC sao cho EFM = 90'.

Chon h$ toa do Oxy sao cho dlnh D trirng v8i g6c 0, c8c di6mC v8 A 16n l ~ q thm tren Ox va Oy (h.2.24).Ta c6 D = (0 ;O), C(a; O),

A = (O ; a ) , B = (a ; a). 'Theo gid thi& ta tim dw c toa do A/& BGiH sB M = (a ; y), ta c&n tim tung doy &a di6m M.Tac6 F E = -.-- (;,;I

--E F I FM e E.FM = 0 Hinh 2.24

- 11-Ta suy ra vecta BM = -BC .6

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52. T~CH O H m G UA HAi VECTU 59

V i du 4. X6c dinh g6c gigs c6c c ap vectd sau d2y :a) = ( O ; - 5 ) ; c=(&;l) ;

- - -c) Ta nh$n thgy vectu p = n -m = ( a m = p = 2 . T a1-1 1-1h8y tinh g6c giira vA i .

Hinh 2.25+ d - . - 41 = 1 ' = 2 n6n vecta n = m + p tao vd i m vA p cPc g6c- - - w ebhng nhau nghia 18 ( m , n ) = 15' (vectu n = m + p c6 do dAi

bhng d~lmgh6o ciia hinh thoi e6 canh 121 1 = 1 = 2 vB c6- -Nhon xkt. Ne"u ta tinh trqc tigp g6c (m,n ) tir cbng that

th i sE gap nhi6u rdc r6i trong khi thgc hi$n cAc phCp tinh.

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60 Ch~rong I. ~ C H H ~ G~ , AA1VECTCJ...

11 . Cho tam gidc ABC vudng tai A c6 g6c 6 = 60'. Hay t inh cdc

12 . Tam gific d& u ABC c6 c anh bang a v2I c6 dllbng cao AH. H aytin h cbc tich v6 hubng sau day :

--C) AC. H A .

13 . Cho doan th in g AB c6 do d i i 2a v i mat s$ k2. Tim tap hqp cdc--i6m M sao cho M A .M B = k2 .14. Cho tam gidc ABC c6 BC = a, CA = b, AB = c. Tinh tich v6

15. Cho tam gidc ABC c6 g6c 2 = a < 90'. VC ben ng oi i tam gidcdZ cho c5c tam gidc vu6ng can din h A la ABD v& ACE. G oi Mla trung di6 m clja canh BC. ChQng min h rang A M vudng g6cvdi DE.

16. Cho b6n digm phan biet A, B, C, D cirng thuoc mot d u h g--thing, bigt rang IEI= a , l E l = b . Hay t inh AB.CD kh i-AB va ccng hudng v2I ngugc hudng.

17. Cho tam gific ABC c6 BC = a, CA = b, AB = c. VC v6 phia ngoaitam gidc h ai h in h vubng ACEF v21 BCDL. ChQng minh :

18. Cho tam gidc deu ABC c6 canh bang a. Ha y tinh cdc tic h vdhubng sau :

b) AB. 2E- AC)- . - . AC) (AB-AC).(AB + AC) .

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19 . Cho hinh vuang ABCD canh a. Tinh c6c tich v8 h ~ d n gau :

- . - -c) (A6-AC).(2AB +AD).-. + +20. Cho hai vecta a = (2 ; 4) , b = (6 ; 1). Tim vecta x = (u ;v)

4 + 4 4saocho a . x =6vA b . x = 29.-.21. Cho hai vectu a = (4 ; 5), 6 = ( 1 ;2 ) . Tinh c6c tich v 8 hddngi.6 vA (;+$).(;-GI.

22. Tim g6c giQa cdc cap vectd sau ddy :4 4a) a = (4 ; ) v i b = (1 ; 7) ;4 4b) c = (2 ;5) va d = (3 ; 7) ;

23. Cho hai digm A(l ;2) vA B(-3 ; ). Tim digm M(x ; y) sao choMA I B vA MA = AB.

24. Tim toa do true tam H(x ;y) clia tam gi ic ABC bigt cdc digmA(2 ; 1, B(-1 ; ) ,C(-2 ; 3).

25. Cho tam gi6c ABC c6 AB = Scm, BC = 7cm, CA = 8cm. Tinh-.-AB.AC v i inh g6c A.- C I A26. Cho tam gi6c ABC c6 ba g6c A,B,C d6u la g6c nhon. Dddng

trbn tdm 0 b6n kinh R ngoai tigp tam gidc d6. ChUng minh rang :

3b) cos 2A + cos 28 + cos 2C 2 --.2

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c6 dubng cao AH = h,B C = a , C A = b , A C = c .I AGoi B H = c', CH = b' .Hay di&n vao c6c 6 trdngsau d i y de" dwc c i c h@thirc dljn g :

sinC = cosB = -a

Ban hag thll xern :

I= - absinC2= di@ n ich AABC.

Chia cAc so" nay cho1- bc ta c6 :2

a ' b c-=-=-sinA sin0 sinC '

TAMChirng ta bi6t rhng mat tam gific d ~ u coan toan xac dinh ne"ubi6t ba canh, hogc hai canh vh g6c xen giila, hoac mat canh vahai g6c k&.Vdi ba truhng hap n6u tren so" do cac canh va cAc g6c cbn la icria tam giac hoan tohn xac dinh. Nhu v$y giaa cfic ye"u to" cliamat tam giac c6 mat m6i lien he xac dinh nao 66. Nwhi ta goi66 la cac hg thQc Zupzg rong ta m g iac . Trong ph6n nay chlingta sG nghien c h nhilng he thI3c d6 va cac & - ~ gung cGa chung.1 . DInh lisina) Cho tam giac ABC vuSng tai A c6 BC = a, CA = b, AB = c(h.2.26). Ta dS bigt : A

b bs i n B = - * a = -a sin Ba sinL a

n Cb c D-o66 a=--sinB sinC ' (1)

Vi A = 90" 6n sin A = 1va t a c6 the" vie"t he thI3c (1)nhu sau :a b c-=-=-sinA sinB sinC '

Hun n3a vi BC = a = 2R 18 duhng kinh duhng trbn ngoai tie"ptam giAc vu6ng ABC n6n :a-- - b-- - C-- - 2RsinA sinB sinCHe thtlc nay d q c goi 18 dinh l i sin trong tam giBc hay d ~ u cgoi t h t 18 djnh l i sin.

p b) Dinh li . Vdi ta m gi5c ABC ba"t kki, t i so" @Cia m$t canh vAj sin c i a g6c dbi dien v6i canh d6 lu6n lu6n bang d ~ a n g inh

C c i a d ~ i r n grbn ngoai tigp ta m gihc.U a b . c-=-=- = 2R.I sinA sinB sinCT a d3y hay suy ra dinh l isin.

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53. CGC HE THUC LLJQNG TRONG TAM GIAC VA GIAI TAM GIAC 63

C H ~ N G INHd myc a) dinh li dii d ~ d c h h g minh trong truang hdp g6c-AC= 90". Ta c8n c h ~ g inh dinh li trong tnlang hup g6cCIBAC kh6c 90".-6u g6c BAC nhon ta vG d&ng kinh BD ciia duang trbnngoai ti6p tam giAc. Khi 66 vi tam giAc BCD vu6ng tai C n6nta c6 :

BC = BD sinD hay a = 2R sinD. (h.2.27a)CC - ATa c6 BAC = BDC v i 66 1h hai g6c nai tisp c h g c h h cung BC .

aDo 66 : a = 2R sinA hay -- - 2R.sinACINe"u g6c BAC th ta cfing vE d u h g kinh BD cGa d d h g trbnngoai tie"p tam gidc ABC. TQ gidc ABDC nai tigp d d h g trbnd u h g kinh BD n6n 6 = 180' - . Do 66 sinD = sin(180- A).Ta cfing c6 BC = BD sinD = BD sin(180- A) = BD sinA.

aVay a = 2R sinA hay- 2R.slnAb cCic dhng thQc -- - 2R V B -- - 2R duqc c h h g minhsin B sin C

He quci. V6i moi tam giAc ABC nai tigp duang trbn bAn kinhR, ta c6 :

a = 2R sinA ,b = 2R sinB , c = 2R sinC.c) V i du. TCf hai v i t r i A, B cAch nhau 25m v5 cirng thuoc motba sBng hay do khoang cdch ti l v i t r i A v i v i tr i B td i vi t r i C d-$n k ia dbng siing, bigt rsng BAC = 110" v i ABC= 20"(h.2.28).

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64 Chdung II. T/CH vo H ~ G~IAA1VECTIl... .,

A BHinh 2.?*-a c6 ACB = 180" - 110" + 20") = 50"

Theo djnh li sin ta c6 :sin110' - sin 20' sin 50'- -a b 25

25. sin 110'Vey a = - 30,7 (m)sin 50'25. sin 20'b = z 11,2 (m).sin 50'

a) BAi tohn. Cho tam gihc ABC bi6t AB = c, AC = b vA bigtg6c A. H5y tinh canh BC = a (h.2.29).

-2 - - -2 -2 --C = (AC -A B ) ~ AC + AB - 2AC.AB

Hinh 2.29

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53. CA C HETHUC L W G RONG T A M GIAC VA ~ l d l AM GIAC 65

1- AC = 5. Hay tinh ca nh BC.L

f I a) khi g6c i 90" ;[ I b) khi g6c = 60".

theo gi6 tri clja he thOcb2 + c2 - a2 IA m ot s6 duong,hoac sd im, hoac bang 0,ha y suy ra tinh ch5t cca g6cA (tO, nhon, vudng).

Tif ke"t q& cfia bhi todn t ren ta suy ra dinh li sau diiy :m7 h) Dinh li. Trong mat tam giic ba"t kki, b in h p h ~ ~ n gat canhj I~j biing tbng binh phdung hai canh cbn lai trllr hai Idn tich hai1 canh kia nh2n vdi c6sin cSa g6c giila hai canh d6 .1

Trong moi tam gidc ABC, vdi BC = a, CA = b;AB = c ta lu6n c6 :a2= b2+ c2- 2bc COSAb2 = a2 + c2 -2ac cosBc2= a2 + b2- 2ab cosC.

t o5Cho tam gidc ABC w b n g taiA, c6 BC = a, CA = b,A 6 = c.

I Hdy dua "20 cbng t h O I2 2(b2+ cZ)- a2m a= 4d6 tinh da d i i rn,.

Vdi moi tam gi6c ABC, vdi BC = a, CA = b, AB = c ta luan c6 :

2 2 2a +c - bcosB =

C) Tinh do dai duirng trung tuye"n ciia tam gidcCho tam gidc ABC c6 BC = a, CA = b, AB = c. Goi ma ,mb ,m,la do dAi chc d u h g trung tuye"n ldn l m t vG tif cbc dinh A, B, Cclia tam gibc. Khi d6 ta c6 : A

2 2(b2 + c2)- a2m =a 42 2(a2+c2)-b2m,, =

42 2(a2+b2)-c2m =C 4 B M C

Hinh 2.30That v@ygoi M la trung digm ciia canh BC. Ap dung dinh lic6sin vao tam gi6c AMB ta c6 (h.2.30) :

22 am2 = ~ ~ + ( ~ ) ~ - 2 c . ~ c o s B = c-- ac cosB2 2 2a +c - bVi cos B = ni2n t a c6 :2ac

2 2 a ac. a 2 + c 2 - b 2 - 2(b2+c2)-a2ma = +--4 2ac 4

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g @.Tam gi6c ABC c6 di@ntlch S- N& k6o dhi cqnhAB mot doan AB' = 2A Bva kc50 dhi canh AC motdoan AC' = 3AC. G Q ~' 13di@n lch tam gidc AB'C'.

S'Hay x6c dinh t1 s6 -5

C h h g minh t w n g tg ta c6 :

b) vi du. Ch o tam g iic ABC c6 a = BC = 5cm, c = AB = 8cmIvA g6c 6 = 77'. Tinh canh AC, c6c g6c A, C cda tam gi6cvA d&ng trung tuye"n mb = BM cda tam gi6c d6 (h.2.3 1).G I ~ I B

Theo dinh li casin.'ta c6 :A C ~ b2= a2+ c2- 2ac cosB

= 52+ s22.5.8.cos77O= 89 - 80 cos 77'.

A C ~ b2 71. /A CV$y AC - f i 8,4 (cm). Hinh 2.31Mu6n tinh g6c 2 t a dDng djnh li sin :sin A sin B-= - a. sin B=,sinA =-a b b5. sin 77'Do 66 sinA = f i - 0,5782

e - 180' - 77' + 35'19') - 67'41 ' .dung cbng thirc :

3. Cdng thljlc tinh dien tich tam giacCho tam gidc ABC c6 BC = a, CA = b, AB = c. Goi R vh r l&nl%t lh ban kinh d&ng trbn ngoai tie"p, nai tizp tam gidc va

a + b + cP = 1B nfia chu vi tam giAc.2Ta c6 cAc cang thirc tinh di$n tich S clia tam giac ABC sau day :a) Cac cbng thilc :

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53.CAC HE THUCLWNG TRONG TAM GIAC VA ~ i d i AM GIAC 67

abc2 ) S =-RTa chlihg minh cdc cdng th3c trCn :

A

Tim die,n tich clia mai tamf ~ l inh 2.32 11)Tam giBc ABC c6 di$n tich S = -ah, v6i ha = AH. Vdi g6c2nhon, th hay vudng ta d&uc6 :AH = AC sinC = b sinC (h.2.32)=

!- ) Gpi I 1P tiim d ~lrng rbn niji tiirp tam giBe ABC. Ki hi$"I SaBC a diijn tich tam giBc ABC. BBn kinh r c6a dlllrng trbnniji tigp tam gidc la d~lrngao xu& phAt til 63nh I cba cac tamt giBc IBC, ICA, IAB. Ta c6 (h.2.33) :1Do 66 S = - b sinC.2IIHinh 2.33

b)

10c )

5

1 1CBc cdng thGc S = -ac sinB vP S = - c sinA d~ u c h h g2 2minh t m g g.

C C2) Theo djnh li sin ta c6- 2R hay sinC =-in C 2R '1 1 c abcTa c6 S = -ab sinC = - a b =-2 2R 4R '

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68 Chlrung II. *H @ H U I ~ GC~IAA1VECTII...

2 2 2b + c - aTheo dinh l i cosin ta c6 cosA = . Do 66 :2bc2 2

1+ cosA = (b + c ) - a - ( b + c - a ) ( b + c + a )2bc 2bc1Vay S2= -(a + b + C) (a + b - C) (b + c - a) (C+ a - b).16

c + a - b = 2(p - b).Ta suy ra S2= p(p - a) (p - b) (p - c)hay S = Jp(p - a)(p- b)(p- c) .b) V i du. Cho tam gi6c ABC c6 AC = 7cm,.AB = 5cm. Bi6t

3cosA = -, hZy t inh c an h BC, die n t ich S cGa tam gist,5 .duting cao ha v i b6n k inh R c i r a d ~ t i n gr6n ngoa i ti6ptam gidc.

4 iia gpi tam giAc c6I do d i i ba canh li ba so" InguyPn liPn tig p v i c6 dientich bang mot sd nguyCn litam g i i c He-r6ng. Cdc tamgidc d6 c6 do da'i c6c canhnhu sa u :

- Theo dinh l i cdsin ta c6 (h.2.34) : 3B C ~ a2= b2 + c2- 2bc cosA = 7% 5' - 2.7.5.- = 32.5

Hinh 2.34

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43 inA = -, vi sinA > 0.5

a- Theo dinh li sin ta c6- 2R n6nsin A4. ~ i 6 iam giac va irng dunga) GiHi tam giacGi6i tam giac la tinh cac canh va cac goc ciia tam giac dgatr6n mat so" y6u to" cho tntac. Mu& gihi tam giac, t a c6n timhi6u giii thie"t vh k6't luan ciia bhi toan d6 lga chon cac he th13chu ng thich hap (49dduc n6u trong dinh li sin, djnh li c6sin vac6c c6ng thoc tinh di$n tich tam giac.Vi dl! 1. Cho tam gi5c ABC bigt canh AC = b = 47cm vA cdc

g6c = 48', = 57'. Hay tinh i , va c.

Ta c6 = 180' - (48" + 57') = 75'.Theo djnh li sin ta c6 :

a-- - b (h.2.35)sinA sinB A b=47cm CbsinA 47.sin4a0 _ 36(cm)j =-- -sinB ,qin7s0 Hinh 2.35

C-- b-- b sinC 4 7 . sin 57'-c=-- = 41.sinC sinB sin B sin750V i du 2. Cho tam gi6c ABC c6 c6c canh a = 49cm, b = 26cm

vA g6c = 47'. Tinh canh c vA cdc g6c i, h.2.36).

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70 Chcrdng II. T~CH Q H U ~ GGA HA1VECTU.. .

Hinh 2.36Theo djnh li casin t

c2 = a2+ b2 - 2ak - - - -c2 = 4g2 + 262- 2.49.26.cos47'c2 = 2401 + 676 - 1.737,73 = 1.339,27.

V$y c = Jm36,6 (crn).V$y x la g6c th vh ta tinh d ~ u c = 101,42'.V i dl! 3. Cho tam gidc ABC c6 cdc canh a = 24cm, b = 13cm,

c = 15cm. Tlnh g6c x , dien tich S cda tam gidc v i cdcbAn kinh R, r c i a cAc dddng trbn ngoai tie"p, nQi tigptam gidc.

Theo dinh li c6sin ta c6 (h.2.37) :

Hinh 2.37Nhu v@yA 18 g6c th vB ta tinh d ~ q c = 117'49'.

Goi S 1h dien tich tam.giacABC ta c6

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53.CAC HE THUC UONG TRONG T AM GIAC VA GIAI T A M GIAC 71

&/Hinh 2.38

GIAIAp dung dinh li sin vao tam gihcABD ta c6 :

AD AB-- -- = A D = AB.sin BsinB sinD sinDT a c 6 : X=E+E=E=X-g 63"-48"= 15".

abc abcAp dung ciing thirc S =- a c6 R =- 13,64 (cm).4R 4sS 24+ 1 3 ~ 1 5

Theo ciing thirc S = pr ta c6 r = -. Vi p = = 26P 28575 - 3,3 (cm).Cn r =-26

Chri j . Ta c6 the" tinh dien tich S cia tam gihc bhng cang thircHC-rang nhu sau :

Trong truang hup cu the" nQy,viec tinh di$n tich bhng cangthirc HC-r6ng cho ta ke"t qu& chinh x8c hun dung cang thirc

1S = - c sinA.2b) Ong dung vho viec do dac trong thgc ti$B i i t o i n 1. Mu 6n do chi& cao C D = h cda rnqr c hi t hd p m ita kh6 ng the" de"n d ~ w cdm C cda chdn thdp. Trong matphAng dQng chOa ch i& cao C D cda thdp ta chon hai die"m A,B tr$n mat da"t sao cho ba d ik m A, B, C thsng hhng. Gi5 st3 taCIdo d w c khodng cdch AB = 24cm vA cdc g6c CAD = 63" ,(CCBD = 48". H a y tinh chi& cao h clja th5p (h.2.38).

I h = CD = AD sinA = 68,91.0,891 z 61,4 (m).

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Do 66 AD = 24. s in 48' 24.0,7431z = 68,91 (m).sin 15' 0,2588BPi to611 2. KhoAng c6ch tC A de"n Bkh6ng the" do trgc tigp d w c v i phaiqua mat c6i h8 nu &~. gUtfi ta chonmot di6m C tao nen tam gihc ABCc6 BC = loom, AC = 80m v& g6cCIACB = 48'. Tinh khoAng cich AB(h.2.39). c

Hinh 2.39

Ap dung dinh li c8sin d6'i vdi tam gihc ABC, ta c6 :

4 4

B i i todn 3. Hai luc ddng quy P,Q tao v6i nhau mot g6ca = 50'. Tinh hup lgc v i dnh gbc tao bdi R vdi p v ~vdi 6 i6t r5ng ? = 1OON, a = 200N.

GIA- - - - - -a vi5 Pd = P,OQ = Q vh OR = OP+OQ. Khi 66 POQ = 50".X6t hinh binh hAnh OPRQ (h.2.40), ta c6-PR = 180"- 50" = 130".A~ ung dinh li c8sin d6i vdi tam gihc OPR ta c6 :

Hinh 2.40

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5 3 . CAC HE THOC LUQNGTRUNG T A M GIAC VA ~ l d l A M GIAC 'i3- p2+ 0R2 -P R ~ l oo2 + 275i2 - Y ^ ^ ~cos POR = -2.OP.OR 2.100.275

7 -a suy ra POR 2: 34" vA QOR = 50' - 34' - 16" .5.Cac dung bai tap cd b8nDqng 1. Tinh mat s6 ycu to" cua tam giac the0 mot so" y6u to"cho trudc (trong 66 c6 it nhgt 11B mat canh).Phuung phap .- S3 dung trgc tigp djnh li sin v&dinh li cbsin.- Chon c5c he thilc hung thich hqp d6i vdi tam gi6c de" tinhmat s$ ye"u to" trung gian c6n thigt vvB sau 66 tinh cac y6u to"c6n tinh.- Sd dung dinh li "Tdng cAc g6c c3a mat tam giac bhng 180""V i du 1. Cho tam gi5c ABC c6 a = 34 . = 45'. ? = 64'. Tinh

g6c A v i c5c canh b, c cda tam gi5c d6 (h.2.41).

= 180"- (45" + 64") = 71".Theo dinh li s in ta c6 : / k-- b c--=-sin A sin B sinC 5 64.)\

B a = 34 Ca sinB 34.sin 45' 2: 25,427-o d6 : b = -- Hinh 2.41sin A ,in71Oa sin C 34. sin 64'-= -- 2: 32,32sinA sin71"

Chri 9. Ta c6 the" tinh canh c the0 cdng thilc : .b sin C 25,427. sin 64'-= --sinB sin 45'

V i dl? 2. Cho tam gidc ABC c6 a = 50, b = 27 , = 48'. Tinhha i g6c A, B va canh C.

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74 Cht fmg II. T~CH o HUT ~G ~ AA1 VECT(7.. .

Vgy c = , = 37,714. Hinh 2.42

Dhng m6y tinh b6 tlii ta tinh duuc x = 99'51'.I

Tir d6 ta tinh d q c % = 180' - E + x)1f j = 180' - (48' + 99'51') = 42'9'. iiV i du 3. Cho tam gi5c ABC c6 a = 48, b = 26 , c = 30. Tinh 1

C ~ C6c A, B, C. 1

Theo he qud cSa dinh li cSsin ta c6 :A

Ta tinh duuc x r 117'49'.a b--heo dinh li sin ta c6-sinA sinB '

b sinA 26.sin 117'49 '-o d6 sinB =- = 0,4791.a 48FHinh 2.43

Tam giac ABC c6 canh AC ngdn nhgt n6n g6c B nhon.Ta tinh duuc = 28'37'. h

Dqng 2. C h k g minh c6 c h e that v6 m6i quan h e giGn c;ic y6ut 6 cua met tam giic.

Dhng c8c he th3c cca bdn de"bign d6i ve"nay thanh ve"kia, hoecchlihg minh hai ve" chng b5ng mat bigu thoc thQ ba, hoec

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53.CAC HE TH& LUM\IG TRONG TAM GIAC VA GIAI TAM GIAC 7.5

c h h g minh h$ thGc c6n c h h g minh tumg d ~ u n gdi mat h$thGc dd bigt la d6ng. Khi c h h g minh c6n khai thac cac gihthigt vh kgt luan de" tim duuc cAc he thGc thich hap 1Bm trunggian cho qua trinh bign d6i.V i du 1. Goi G la trong tdm c i a tam g i i c ABC v i M I2 mat

d ikm tujl 9. ChlSIng m inh r i n g :a) MA^ + M B2+ M c 2= G A ~ GB' + GC' + ~ G M ~

9b) ma + mb + m, 2 -R, trong d6 m a , mb , m, Id n Idd t2I & do d i i c6c trung tuye"n Ong vdi cAc can h a, b, ccGa tam gi6c v i R I i b i n k inh dddng tr6n ngoai t igptam gi6c ABC.

- - - -GA2 + G B ~ GC2 + ~ G M ~2GM.(GA + G B + GC)- - -G A 2 + G B ~ G c 2 + 3 ~ ~ ~ ( 1 ) ( v iA + G B + G C = ~ ) .b) Theo h$ th3c (1) a lu6n lu6n c6

3 3 3Met khAc ma = -GA, mb = - GB, m, = - GC nen :2 2 2

Thay M b8ng tam 0 cba d&ng trbn ngoai ti6'p ta m giac taduuc :

9hay m , + m b + m c I - R .2V i du 2. Cho tam gi6c ABC vudng tai A. Goi m a, mb , m, I$nl ~ ta do dAi c6c d~ivng rung tuye"n xu& p h i t tCt cdc

dlnh A, B, C cfia tam gi6c. ChUng m inh r i n g :

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76 ChUmg II. T ~ H H ~ G~ AA1VECTO... IG I ~ I

Theo c6ng that tinh&j hi 6 d m g tru ng tuye"n ciia tam giac ta cb : t -i1Vi ta m g i6c vu8ng ta i A n6n t a c6 a2 = b2 + c2 v a ma = - a2

V i dl! 3. Cho S IA dien tich cua tam giic ABC. ChUng minh : I

1Ta b ig t r i n g S = -AB.AC sinA2

1 -2 -2 --= -[AB AC - A B . A C ) ~4V i dr! 4. Cho tam gi6c ABC c6 BC = a, CA = b, AB = c vA r IA

bAn kinh du'irng trbn ne i tie"p. ChUng minh r%ng 1tanA c2+a2-b2 ,-) -- 2 2 2 'tanB c + b -a

B Casin-sin-b) r = 2 2A .COS-2

a) Theo d j n h li s in v a chsin t a c6 :-in A - 2R - at a n A =- -- 2bcCOSA b 2 + c 2 - a 2 2 ~ ' ~ ~ + ~ ~ -

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1 53.CAC HE T H l k LCfONG TFiONG TAM GlAC Vi GlAl TAM GIAC 77-- - - - -b-inB - 2R - b- - 2actanB =-COSB a 2 + c 2 - b 2 2 R ' a 2 + C 2 - b 2

2 2 2t an A a + c - b-o 66 ta suy ra- 2 2 2 't a n B b + c - ab) Goi I 1h tAm d ~ a n grbn nai tigp tam giAc ABC. Cdc canhBC, CA, AB tigp xlic vdi dlrfing trbn nay l i n b u t tai M, N, D(h.2.44).Ap dung dinh li sin d6i vdi tam gidc BIC, ta c6 :

-- - ~~ -- -sinBIC sinsin-2 +- - BCB + C -- A (1)sin- os-2 2

Hinh 2.44MGt khic trong tam giAc vuang IMB t a c6 :

. LIB =- 'sin-2Tir (1)vh (2) ta c6 : 1 il- -B Csin -.sin - - A 'COS-2 2

B Ca.sin- in-Do d6 : r = 2 2Acos0

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78 Chllang II. T/CH voHU%GCUA HA1VECTII.. .

V i du 5 . Cho tam giAc ABC c6 A = 60, canh b = 12,C = 6. H % y in h :a) Canh a v2 g6c B clja tam gidc ABC ;b) Dien t ich S clja tam gidc ABC ;C) B5n kinh R clia dllirng trbn ngoai tig p tam giac ABC ;d) 38 dAi d ~ d n gao BH vA dildng trung tuygn A M cGa

tam gi ic ., ' GIAI

a) Theo dinh li c6sin ta c6 (h.2.45):

b) Dien tich S ciia tam giac duuc tinh the0 c8ng thQc :

-1 f i= -.12.6.-= 18&.2 2

INh@nxkt . Vi g6c B = 90" In6n ta c6 the" tinh dien 60'tich tam gihc the0 c6ng A H b = 1 2 6thBc :

1S = -.AB.BC = -.6.- 12&-18fi. Hinh 2.452 2 2

abc 6&.12.6-) T ~ c ~ s = * , ~ o ~ ~ R =- = 6.4R 4s 4.18.ANh&n xk t . Vi tam gi6c ABC vu8ng h i B nen bhn kinh d~irng

1trbn ngoai ti6p tam giac bhng - canh huy6n AC. Ta c62

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Dulrng trung tuyG'n AM c6 do dai duqc ti nh the0 c6ng thdc

Darlg 3. Gi5 i tam gi5c v,i v s n d6 do dac .l'h rltrng ph cip- Tam giac ABC th ~ l rng uuc cho dudi ba dang chu y6u : bi2;tmot canh va hai g6c k6 (g, c, g), bi6t mot g6c va hai canh kgg6c do (c, g, c), va bie't ba canh (c, c, c).- De" tim cac yku to" cbn lai cua tam giac n wdi ta thuirng dungcac djnh li sin, dinh li &sin, dinh li : "Tdng ba g6c cha m a ttam giAc bhng 180"" va dac biet c6 the" dhng chc he thdc lw ngtrong tam giac hoac cac cang thUc tinh dien tich tam giac.V i du 1 . Cho tam gidc ABC bigt toa do cdc di6m A(4& ; 1 1,

B(0 ; 3), C ( B& ;3). Tinh cdc canh v i cdc g6c clja tamgi6c ABC d6.

Nhu vey tam gi6c ABC c6 BC = 8& , CA = 8, AB = 8.

Vgy A = 120". Vi AB = AC n&n am giac ABC c6

V i du 2. Ngllili ta mu6n do khoang c5ch tir hai vi tri A vh B c6chnhau 500m 6 ben nay bd sang tdi vi tr i C 6 bi2n kia bd-ang. Ho do d l l ~ c6c CAB = 87", g6c CBA = 62". Hayneu c6ch tinh cAc khoing c6ch AC v21BC (h.2.46).

a b--- CTheo dinh li sin :- -- ta c6 :sinA sin B sin C

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80 c h ~ m g1 . T~CH o H U ~ I G C ~ , AAI VECTCJ.. I'

Hinh 2.46 ic. sinB 500. sin 62"-C = b = -- - 857(m)sinC sin 31"c. sinA 500. sin 87'-C = a = -- = 969(m).sinC sin 31"

V i du 3. Mu6n bie"t do cao clja thip cao nhst the" gidi 6Toronto thuoc Canada (c6 t2n goi I i "Canada's NationalTower") ngddi ta ti& h inh nhd sau :Goi S I i dlnh thdp v i C I i am c6a chan th5p. Vi khbngdo dduc ti, chdn C clja thip, ngdtfi ta lgy mat digm An i m tr@ndddng n i m ngang vdi chAn C clja th ip v i do :dddc g6c SAC = 70". Tr@n clZlng thing CA ngddi ta IAy .mat digm B c6 khoAng cdch AB = 118m vA do dddcg6c SBC = 60". Ti, d6 ngddi ta tinh dw c chi& cao cljathdp (h.2.47). HGy cho bie"t ho d2 tinh chi& cao naynhd the" nio.

Mu6n tinh do cao h = SC ta cAn Sbi6t d q c do dhi canh huy&n SAcGa tam gihc SAC vu6ng tai C vac6 g6c = 70". X6t tam giAc SABC-ta c6 ASB = 70"- 60" = 10".Ap dung dinh li sin d6i vdi tamgihc SAB ta c6 :

AB-- - SAsin 10' sin 60"Hinh 2.47

Do 66 SA = AB.sin 60" - 118.0,86602- = 588,5(m).sin 10" 0,17365

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Vi sin 70' = ta suy ra SC = SA.sin70SA '= 588,5.0,9:

V(iy chi& cao SC cia the"g i b i lh 553m.

27:Cho tam gidc ABC b ig t :a) =90, 6 = 5 8 ' , a = 7 2 c m . T i n h e , , c ;

Ab) = 90, = 48', a = 20cm. Tinh C, a, c.28. Tinh gbc Idn nhgt c i a tam gidc. ABC vA d i@ n ich clja tam gi%c

d6 trong m 8i trcldng hgp sau :a) a= 40 c m , b = 1 3c m , c = 3 7 c m ;b ) a = 1 3 c m , b = 5 c m , c = 1 2 c m .

329. Cho tam gidc ABC b igt cosA = -, b = 5, c = 7.5Tinh a, SAABC , R V A r.

30. Tinh cdc g6c x , v i d d n g cao ha, bdn kinh dl ll rng trbn ngoait igp R, noi tiG'p r clja tam gidc ABC biG't :a = & , b = 2 , c = JS +I.

31. Goi G la trong t;im clja tam giAc ABC. ChQ ng m inh r?ing.:

32. ChQng rninh r3ng trong tam gidc ABC, ta cb :a) a = b.cosC + c.cosB ;b) sinA = sinB cosC + sinC cosB.

33. Tam gidc ABC c6 chu vi 2p = a + b + c. ChQng m inh rsng. 2sinA = ,-Jp(p -a)(p-b)(p- ) .bc

34. Goi ma , m b, m, I8n l w t Ii cdc trung tuye"n Qngvdi cdc canh a,b, c crja tam gidc ABC.a ) T i n h m a b i e " t a = 2 6 , b = 1 8 , c = 1 6 ;