Hinh Hoc 10 - C2

HA1 VECTC) VA '&NG DUNG

6 1dp 9 ta

0, Cho tam gidc ABC vu8ng I tai A va c6 BC = 2a, CA = a. H a y tinh c6c 11 r6 bong gidc rau ddy :

I

I

di3 bigt dinh nghia ti so" ldung giBc clia g6c nhon a = xOy nhd sau :

La"y mat diem M b6t ki tr6n tia Oy sao cho M khbng trhng v6i 0 r6i vE MP vubng g6c vdi tia Ox. Ta c6 tam giAc OMP vu6ng tai P. Ta goi : sins =

MP canh dbi -- OM canh huydn

OP canh k6 cosa = -OM canh huykn MP canh dbi t a n g a = -= OP canh k& 0 OP canh kd cotang a = -= MP canh d6i'

/

; P

Hinh 2.7

CBc tl so" nhy kh6ng phu thuac vho tri cSa diem M tr6n Oy (h.2.1). B&i v@yta c6 thz chon di6m M sao cho OM = 1 vh t a c6 : s i n a = MP , cosa = OP

Ta c6 the m& rang khdi ni$m ti sd Z~qnggicic czia gdc nhon a thhnh khBi ni$m gia trt Iuung giac czia mot g6c a vdi ' 0 Ia 1180' bang dinh nghia sau dgy :

I

/ I.Dinh nghia. Cho h@tog do Oxy vdi c6c didm A(-, ; O),

ij !i

B ( l ; 0) v i C(0 ; 1). Ta goi n8a d&ng trbn dllirng kinh AB vA di qua didm C la naa d&ng trdn don vj (c6 b6n kinh bang 1 ).

Hinh 2.2

Vdi m6i g6c a ( 0' < a1 1 8 0 ~ ta ) x6c dinh d ~ v cmat di6m M

~ sao cho MOx (h.2.2).

tren n8a d u h g trbn d m

:

=a

GiA s 3 di6m M c6 toa da (xo ; yo). Khi 66 t a dinh nghia : - Tung do yo ciia di6m M goi lh sin cda gdc a v8 duuc ki hi$u lh sin a = yo.

- Hohnh 66 xo ciia 6ie"m M goi lh c6sin cda gdc a vA dwc ki hi$u 18 cos a = xo. v

Yo vdi xo - TI sd -

#

0 goi 18 tang clia gdc a v8 dmc ki hieu

Xo Yo ( a 18 tan a = -

;t

90').

Xo

Xo vdi yo - Ti sd -

#

0 goi lh cbtang cda gdc a vh duuc ki hi&

Yo

CBc sb sin a , cos a , tan a , cot a duuc goi lh cdc gid tri l r l ~ n g giac cda gdc a.

, Cho a vA

p lb

hai g6c

tO trong d6 a c

V i du. Tim c i c gi6 trj lllqng gihc clja g6c 135".

P.

Hay so sdnh : a) sin a vA sin p ;

b) cos a va cos f3

.

Ta 1a"y di6m M trCn naa d&ng trbn d m vj sao cho MOx = 135" (h.2.3).

-

Khi 66 suy ra MOy = 45' v&digm M c6 toa do lh :

Chllmg II. T/CH

30

Vey sin 135" =

HIJ~%JG C~IA HA1 VECTI3.

J5 , cos 135" = --J5 , tan 135" = -1, cot 135" = -1. 2

2

Hinh 2.3

2. Gia tr[ Itfdng giac c6a hai goc bu nhau a) Dinh / j . Hai g6c bC nhau c6 sin b%ngnhau vA c6 c8sin dbi nhau.

i

sina = sin(180° - a ) cos a = -cos(l 80" - a ).

C H ~ N GMINH _4

-

LiXy di6m M tr6n n3a d&ng trbn d m vj sao cho MOx = a r8i

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

lii'y di6m M' d6i xxlihg v6i M qua truc Oy thi xOM' c6 so" do biing 180' - a (h.2.4). R6 r b g la hai 6ie"m M vh M' c6 tung do b h g nhau cbn hohnh do clia chung thi 66i nhau. Do 66 sin a = sin(180° - a ) cos a = -cos(180°

- a)

Hinh 2.4

51. GIA TRI LUONG GIAC CUA MOT GOC

31

I b) HS. qud. Hai g6c bO

la/

nhau c6 tang dbi nhau v i chtang dbi

nhau. tan a = -tan(180°

7

cota = -cot(180°

- a) -

a).

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

sin 150" = sin 30" =

1 fi ; cos 150" = -COS 30" = -- .

2

2 '

tan 150" = -tan 30" = --J3 ; cot 150" = -cot 30" = - & . 3

3. Gio trj lwng giac c6a mat sd goc doc biet

0°

30"

45"

60"

90'

180"

sin a

O

1 -

2

& 2

45

1

0

cos a

1

45 -

J5 -

-1

2

2

0

-1

tan a

O

x

1

fi

II

0

cot a

II

43

1

1 -

O

II

2 1

2

J3

Yo - 1 18 khang xac dinh vA Ta c6 tan 90' = xo O 1 cot o0 = X0 - - cGng kh8ng xdc dinh n8n trong bHng lugng Yo O gidc ciia c6c g6c dac biet ta dcng ki hi& "(I" de" chi giA tri Chri

J?.

kh8ng xdc dinh 66. Mat so" g6c dac bi$t khdc nhu cdc g6c 120°, 135"' 150" d&u c6 cAc gid t~ 1-g giAc d w c suy ra til cac giA tA 1Uqng gi6c ciia CAC g6c d$c biet dii cho.

Ch~ung11.

32

T/CH ~6 H

~ C ~G HA1 A VECTU.. .

V i du. sin 120" = sin(180° - 120") = sin 60" =

J3 2

4. Cac he t h k giila cac gia tri lwng giac cca mot goc

Ta ludn lubn c6 :

sinus1

vi cos a 5 1 .

sn

Dinh 11'. Vdi moi g6c a ta d8u c6 :

0i

I

a) N

* 0 thi tan a

~ U cosa

b) N& sina

#

=

0 thi cota =

sina

-; cos a

cosa . -, sina

c ) sin2a + cor2a = I.

CHUNGM I N H a) va b) : Til djnh nghia ta c6 : YO (vdi xo tana = Xo

cota =

#

sin a 0) hay tan a = cos a

0 (vdiyo + 0 ) h a y c o t a

Yo

=

cos a

-. sin a

Tren n3a d&ng trbn ddn vi ta l6y di6m M sao cho MOx = a . Goi MI va M2 la hinh chi& cria M lhn l ~ d trkn t Ox va Oy (h.2.5). C)

-

-

Ta c6 s i n a = OM2 va cosa = OM1.

Hinh 2.5

33

5 1. GIA TR! LLQNG GIAC CUA MOT GOC 2. Cuc he thdc khcic

5 Ii !i

Djnh 1;. Nku cosa+O thi 1

I

+ tan2a =T.

cos a

i;

1 N 6 u s i n a t O thi 1 + c 0 t 2 a = ~ . sin a

iiI

CH~NG MINH 2 2 2 sin a - cos a + s i n a -- 1 T a c 6 1 + t a n2 a = 1 + -2 2 cos a cos a cos 2 a

cos 2 a

l + c o t 2 a = l +T sin a

-

sin2 a + c o s2 a

-2

--

1

sin2 a

sin a

3. Ap dung a) Cho bigt tanx

x + cos2 x 1 sinx cos x sin x cos x

s i n x cosx sin + cotx = +-- -

tanx Ne"u tanx

+ cotx = 2, h8y tinh sinx.cosx.

cosx

+ cotx = 2 thi

slnx

2

1 = 2. sin x cos x

1 Ta suy r a sinx cosx = - . 2

2 =sin 2 a Do66A=

2 2 72cot a=-+-- 12

sin a

sin a

1 sin 2 a

2 cot2 a

5. Cac dong bai tep cd b i n Dung 1. De" tinh cAc gia tn bung giac clia mat g6c khi bi6t mat gia t n lIfdng giAc c3a g6c 66. Phuungphdp :

-

- Dga vAo dinh nghIa, chn tim tung dij yo va hohnh do xo c3a di&mM tr6n nila dlllrng trbn d m v j vdi g6c xOM = a vh t 3 66 ta c6 : s i n a = y o ; cosa = xo; t a n a =

- ; cota

=

Xo

Xo -. Yo

- Dga vAo tinh chgt : hai g6c bh nhau c6 sin bhng nhau va c6 casin, tang, c6tang d6i nhau.

- Trong mot so" t m h g hqp c6 the" dga vho ti so" do d8i cAc c g h c h mat tam gihc vudng, cu thd 1A do"i v6i g6c nhon a ta c6 :

sina =

d6i -

tana =

d6i (h.2.6) k&

huyQn

Hinh 2.6

D6i vdi hai g6c phu nhau a vh 90' - a ta c6 : sin(90° - a ) = cos a cos(90° - a ) = sin a

vh ne"u 0 ° < a I 90°thi tan (90'ne"u 0 ' I a < 90°thi cot (90'-

a ) = cota

a )= tana.

- Sir dung he that cos2a + sin2a

= 1 d6 tinh tofin.

1 Vi du 1. Cho cos a = -- . Hay tinh c6c gi6 tri llldng gi5c cbn

3 lai clia g6c a v21 so shnh c6c gi6 tri n i y vdi so" 0.

1=8 . Vi s i n a > 0 nen ta lgy Ta c6 sin2a = 1 - cos2 a = 1 - 9 9 2& sin a s i n a = - vti suy ra tan a = -= : = -2Ji 3 cosa 3

(-i)

1 tana

cots = --

1

- --=--

2fi

.J2 4 '

Vi a 18 g6c tir n6n cAc gi8 tfi luang giac cos a , tan a , cot a d&u18 c8c s6 Zim, chi risng s i n a lh s6 dumg. V i du 2. Cho bigt tana = -2. HZy tinh c3c gi3 tri l~wnggi6c cbn 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) Theo he thclc 1 + tan2a =

1

2 ta c6 cos2a

I

=

1+ tan2 a

cos a

1 =1 -

hay cos2 a =

1+4

5'

1 Tir (1) v h (2)ta suy ra cos a = - -

A'

Theo he thirc tan a = hay s i n a = cota =

sin a cos a

- ta c6 sin a

(-5)

2 245 .(-2) = -= -

45

2

tan a

Ccich I. Vi t a n a = cosa > 0.

'

fi.

Tinh bigu thQc A =

tana =

5

1 1 n&nta c6 cota = --

V i dl! 3. 6i6t tana =

Tir he th&

= cos a .tan a

3sina - cosa sina + cosa

fi

1

> 0 n8n a lh g6c nhon, do 66

7 = 1 + tan2a ta suy ra cos a

sin a n&ns i n a cos a

Thay sin a =

:

= cosa.tana =

43 -.& 3

46

=-.

3

6 vh cos a = f i vho bigu thclc A ta tinh d ~ d c: 3 3

C h h 2. Chia ti3 vA mau cba A cho cos a

#

0 ta c6 :

Vi du 4. Tinh : a) sin23" + sin21So + sin2750+ sin2870

b) cos212'

+ c0s*78~+ cos21 + ~ o s ' 8 9 ~

c) Cho bigt sin 15" =

J 6 -J 2 4

. H5y tinh

cdc gid tri

Imng gi4c cbn lai clja g6c 15".

GIAI a) Chli 9 rhng vi 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.

Dung 2. Chdng minh cac he thdc v6 gia tri l ~ u n ggiac cua mat goc a .

- Dqa

vho dinh nghia gi6 tri ldung gi5c ciia mot g6c a vdi 0 I a 1 180'.

- Dqa vao dinh li : "T6ng ba g6c c6a mot tam giac lu8n lu8n b%ng 180'".

- Sd dung cAc he th~Iccu b6n dude suy r a til dinh nghia n h ~ sin a cos a t a n a = - vdi cos a # 0, cot a = - vdi s i n a ;t 0, sin a

cos a sin a + cos2 a = 1. 2

1

- Ap dung he thl'rc :

+N

1 b cos a + 0 thi 1 + t a n2 a = 7 ; cos a

+ Ngusina +

0 thi 1 + cot2a =

1 2 sin a

- Thqc hien c6c phbp bi6n d6i t ~ n dg ~ n g dlra , he th13c cAn chQng minh vd mat he thirc duuc thaa nhan 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' )

= sin4x + cos4x - sin2x cos2x = 1 - 2sin2x cos2x - sin2x cos2x =1

- 3sin2x cos2x

V i du 2. ChQng minh r%ng:

a)

b)

(*)

dpcm : di6u phdi chOng rninh.

1-sina - cosa C O S ~ l+sina tana - sina sin3 a

1 cosa(l+ cos a) '

(dpcm)

T~CHaH U ~UAG HA1 VECTU.. .

C h m g II.

38

1- sin a - (1- sin a ) ( l + sin a) - 1- sin 2 a a> cos a(1+ sin a) cos a(1+ sin a) cos a

-b)

0

cosy a cosa cos a(l + sin a) 1+ sin a

(dpcm).

sina . sin a -sin a tan a - sin a cos a -

.-.

n

sin5 a

sin' a

sin5 a

1 - 1- cosa -cos a ( l + COS a, cbs a sin2a

m).

V i du 3. Chilng minh r i n g :

2

sina b)'-l+cotu

2 - cosa

=sinacosa

a, ~ + c o s ~ [ , - ( l - c ; a ) ~ ] - l + c o s a ( sin a sin a sina

I,'-

- l+cosa(

.

~+tana

1

s i n a (I-=

+--2cota ssrna

1-2cosa+cos a sin 2 a

2 )

cot

-- 1+ cosa -2 cot 2 a + 2cota sin a

sin a

- 1+ cosa .2cot a(sin a

1 sina

--)cosa sina

- 1+ cosa .2 cot a. 1- cos a - (1+ cos a)(l- cos a) sin a

b) l -

sin2 a

sin a

sin2 a - cos2 a sin 3 a = 1l+cota l+tana sina+cosa = 1-

-

.2 cot a

cos3 a sina+cosa

sin 3 a + cos3 a sin a + cos a 2

2

= 1- (sin a + c o s a - s i n a c o s a ) = sinacosa

(dpcm).

J

51.

39

GIA TRI L m G GIAC CUA M?T GOC

V i du 4. Chl?ng minh chc bi@uthirc sau khbng phu thuoc vho x.

+ (sinx - C O S X;) ~ b) B = sin6x + cos6x + 3sin2x cos2x .

a) A = (sinx + cosx)'

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

+ (sin2x + cos2x) - 2sinx cosx = 2.

b) B = sin6x + cos6x + 3sin2x cos2x =

Vi dl! 5. Cho tam gidc ABC. ChQng minh rhng : a) sinA = sin(B + C) ;

b) cosA = -cos(B

+ C) ;

A+B C c) sin= cos2 2

C)

A+B sin-= 2

.

C 2

cos

Vi du 6. Tam gi5c ABC c6 A6 = AC = a vA c6 g6c nhon = 2 a . Goi AH v3 BK lA cAc d&ng cao ve ta cdc dinh A vA B. a) Tinh do d;ii cdc dddng cao AH vA BK the0 a vA a .

b) Chlllng minh r3ng sin2 a = 2sin a cosa. A

--

GIAI

a) Ta c6 BAH=HAC = a (h.2.7)

AH -- cosa a AH = a cos a AB

a

Hin h 2.7

Chlrong II.

40

T/CH

vC)

HU~NG CGA HA1 MCTU...

b) Goi SAHCla di$n tich t a m ABC, ta c6 : 2SAUC= AH.BC = BK.AC trong 66 BC = 2 B H = 2.a s i n a .

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

(dpc

1 . H i y ph6t bi6u dinh nghTa gi6 tri ldung gidc clja 1n6t g6c a vdi 0" 5 a 5 180". a) GiA sir cho g6c a = 120". Hay cho bigt c6c gi6 tri lclung gidc : sin 120", cos 120", tan 1 2OU, cot 120'.

b) Vdi nhCfng gid tri n i o cila g6c a ta c6 :

- sina v i cosa cirng da"u ? - sin a va cos a khdc ddu ? 2. Goi a la so" do clja g6c MOx tr@nnda dclbng trbn dun vi. Hay x6c dinh VI tri ciia digm M tr@nnrlla dddng trbn dun vi d6 sao cho : 1 a) cosa = - . H i y tinh so" do cGa g6c a d6. 2

b) sina =

1 -.

C) cosa =

--.1

2

2

-

Khi d6 g6c MOx la g6c nhon hay g6c tir !

-

Khi d6 h i y t i n h s6do cira g6c MOx.

d) sina dat gi6 trj Idn nha"t. 3. Hay so sdnh c6c cap gi6 tri lcr~lnggi6c 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 bung gidc clja mat g6c a (0 5 a 5 1 80"), h i y tinh gi6 tri c6c bi6u thOc sau (vdi a, b, c la nhang so" cho trudc).

+ b.cos 0" + c.cos 90" b) B = a.cos 90" + b.sin 90" + c.sin 180" c) C = a2.sin 90" + b2.cos 90" + c2.cos 180" a) A = a.sin 0"

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 ~c6c bi6u thrSlc sau : a) M = 3 - sin2 90"

b) N = sinx

+ 2cos260" - 3 tan2 45" ;

+ cosx khi x bang 0" ; 45" ;60";

C)

P = 2si1ix + cos2x khi x bang 60" ; 45" ; 30" ;

C)

() = sili'x

+ COS'X

khi x bang 1 35" ; 1 80" ; 160"

.

6. Chifng minh c i c hhng d i n g 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. Hay tinh cac canh AB, BC v i gbc B clja tarn giac. 8. Cho tam g i i c ABC vu611g tai

I

(

va ccj canh BC = 12,3,

CA = 31,G. H i y tinh canh AB va c,6c goc /i,6 c ~ i atam gi6c. 9. Chirng minh rang a) Vdi g6c u (0' 5 a 5 90') ta c6 :

sin 190" - a ) = cos a cos (90" - a ) = sina ; b) Vdi g6c a (0" c u I 90") ta c6 :

tan (9U" - a ) = cot a ; C)

Vdi g6c a (0' I

cl<

90') ta c6 :

cot (90" - a 1 = tan a . 10. Chdng minh r i n g

1

2

2

sin acos a

2

=tan a + c o t a + 2 .

Chumg II. ~ C vo H HUI%JG C ~ HA1 A VECTCJ.. .

42

1 . Goc gi3a hai vectd

8

,lg/lja.

-

a

Q,

Cho t a m g i j c AUC vuGng c j n tai A

a

d6u k h l c vectd 6. Til Cho hai vectd v i mat digm 0 ba"t ki ta vP =i v i = g. G6c AOB vdi sb do t i l 0' dgn 180' duuc goi la g6c giGa ha; vecto V A g .

r a) Djn/l

ci, I la trung

1 Ta k i hi& g6c g i h hai vecta a v i i; 18 (a, 6 ).

c1ic"m cda c a n h BC.

a

bl C/,.ri 3. Cdch xdc dinh g6c giaa hai vecta vB 6 theo d/nh nghia n6u tr6n khBng phu thuoc vAo viec chon digm 0. Hun n a a t a c b n c 6 ( ; , G ) = ( g , i). N6'u

(

a, g

)

= 90"thi ta n6i rhng hai vecta

vd.i nhau vh ki hieu C)

a Ig.

a vh $

vudng gdc

V i du. 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 vo hddng c6a hai vectd Trong truilng hap n i o

rich v6 hudnp

;.

6 cO

gia trl cfuong, co gia tri jrn, bang O !

k I

I

a

d&u k h l c vectu 6 a ) Dinh n g h i < ~Cho . hai vecta va la met sb, k i hi& la v6 hu'dng cda hai vecto vd d d ~ xdc c d ~ n hbdi cirng thirc sau :

a

Tr~rirnghup n6u c6 mot trong hai vectu -. ta quy udc a . b = 0.

-

a vh

bhng vectu

6

b ) CIlri 2. Trudc diiy ta d5 bi6t t6ng cua hai vectu la mat vectd, tich ccia mot s6 vdi mot vectu 18 mot vectu. d d i y ta c6 tich v8 hudng cua hai vectd la nzQt sd chd khdng phai la mat vectu. Vi v$y n6n ngu&i ta goi tich 6 6 la tich v6 h ~ d n gc ~ i ahai vectd.

i

Cho tam gi6c d6u ABC c6 canh a va trong tdm G. Tinh c i c tich v6 hlldng sau ddy : c ) V i du.

0, Vdi hai sd thuc J , b bSt ki, ta c6 (a.blL= a'.bL. v b y vdi hai vecta

-.a, b-

b i t ki, t l j n g thUc :

Theo djnh nghia ta co (h.2.9):

I

c6 dung kli61ig ?

--

AB.AC = a.a. cos 60'

1

= -a

2

2

AC.CB = a.a. cos 120' = --a1 2 -7

2

--

a" a 43 GA.BA = -.a. cos 30 = 3 2 d j Di6u kien vudng g6c ciia hai vectu

Tit djnh nghia tich v6 h ~ d n gcua hai vectu thgy r i n g : Ta cluy ~ l d cxcrn vcctu vubng goc vdi rnoi vecw vi

u.a=o.

--

A

A\

Hinh 2.9

vB

ta nhan

- - lit .I61.cos 90' = o ; --. - a . b =Othido a*O,b*6 tasuyra

- N ~ (Ua , b ) = 90" thi a . b = -Ngugclainbutac6

Do 66 ta c6 dinh li : "Ridrr hien cdn u& dii d6' hai vecta a7 t 6 vci 6 + 6 ou6ng gcic u u i nlzau la tich v6 hudng ciia chiing h&ng 0".

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

- coscp . IF^ .l0o11. 4

A=

(h.2.,10)

(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 tdt trong 66

-

vh 00' tinh bhng 60, cbn la m), cp lh g6c gi3a hai vectu c6ng A tinh bhng Jun (vigt tt$t18 J).

IF^. 100'1. cos cp

d

Tmng tofin hgc, giA t* cba bigu thm A =

k& d m v j do dmc gpi 18 tich va h u h g cQahai vectcr

@

"21

kh6ng 00 ' .

f

0'

0 Hinh 2.70

3. Binh phlldng v6 hllang

- -

8

4

-

CH~]Y :

Vdi m ~ doan i thang AB ta lu8n lubn c6 :

A B= ~

4

-

1 - 1 1-1

1-l2 .

Ngu a = b tac6 a . b = a . a = a . a .cosoO= a

z2= [q2

sAB=@.

Khi 66 tich v6 hudng

a. a d@c ki hi& 18 a 2 .

-2

Sd a nhy dduc goi 16 binh p h m g u6 h ~ h cgcLa vectu

i.

Do 66 ta c6 djnh li :

"Binh plzuung v6 h ~ d n gcria mot tlecto b&ng binh ph~crngdo dai cria vecta dd"

4. Cac tinh ch6-t c6a tich v6 hlrCjng Ng&i ta dii c h h g minh duuc cAc tinh chgt sau dAy cba tich v6 hudng : 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) ;

92. T~CHVO H

4 -5

~ CUAG HA1 VECTCY

Tif c8c tinh chgt cclja tich v6 hudng, ta c6 th6 suy ra cAc h$ th3c sau diiy : -2

--

'2

+ - + 2

-2

(i+Gl2=a +2a.b+b - + 2

(1)

(a-b) = a - 2 a . b + b

- - - -

-2

(2)

-2

-2

(a+b).(a-b)=a -b

-2

=la1 -1bl

(3)

5. Cac bai toan a p dung

a

B i i to6n 1. Cho hai v e c t ~ chi&" cda vecta -,*

6

vA

6 bgt

kki, g ~ ib; la hinh

tren d&ng thang chQa vectd

a . ChQng

+ -

minh r i n g : a . b = a . b l

a)

b) Hinh 2.7 7

- -. Ta vG ciic vectd OA = a vh

=

g. Goi B' 1h hinh chi&

tren dlrZIng thhng OA. Ta c$n c h h g minh :

--OA.OB OA.OB1 - Ne"u AOB < 90' ta c6

(h.2.11a) vA b)).

=

:

CI

--.

= 0A.OB.cos AOB = OA.OB .

- Ne"u AOB

C4

> 90" thi :

- --

= OA.OB'.cos AOB = OA.OB .

cclja B

C h m g II. T~CH

46

-

-

- N6u AOB

H W G &A HA1 VECTU...

6

= 90" thi OBI = vB khi d6 --, OA.OB' = OA .O = 0 = OA.OB cos 90" = OA.OB .

--

A-

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 la trung digm ciia doan thhng AB (h.2.12) ta c6 :

-MA.MB = (MO + OA) (MO + OB) -

_

.

A

-

Hinh 2.12

Viiy tep hap cAc digm M lh d&ng trbn tarn 0 , bAn kinh

R=

J m .

B i i toin 3. Cho dddng trbn (0; R) vA dikm M c6 dinh. Mat dddng th3ng A thay dbi lu8n lu6n di qua M cdt dlrirng trbn d6 tai hai dikm A vA B. ChQng minh r i n g :

M A . M B = M O ~ - R ~ = ~ ~( v -d Ri d ~= M O ) . CH~NG MINH

-

VEi d ~ h kinh g BC ciia d ~ & ntrbn g ( 0 ; R). Ta c6 MA 1B hinh chi& c6a MC tren dudng th&ng MB. Theo cbng that hinh chie"u (BBi toftn 1)ta c6 :

/B

M Hinh 2.73

52. T~CHVO H

47

~ CUAG HA1 VECTIT

--- MA. MB MC.MB = (MO + OC).(MO + OB) =

--

Chri .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 vdi

trbn (0;R) vdi T la ti6p

- Khi MT la tidp tuygn cua die"m,ta c6 :

Hinh 2.74

6. ~ i d thilc u toa do cGa tich v6 hdtrng

-. -.

Trong m8t phing toa do (0 ; i , j ) cho hai vectd :

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

c h m g 11.

48

WH~ fHUGNG l C~IA HAI VECT [I...

7. Ong dung c3a tkh vb hlrCIng a) Do dai c6a vectd

a = (al ; a2) d ~ u ctinh theo cdng thtlc :

4

b) G6c gigs hai vectu

0, a) Cho

6

Theo dinh nghia tich v8 h d n g t a c6 :

a = (1 ; -2) ;

= (-1 ; -3). Hay

tinh g6c (

b) Cho

a

= (al ; a2) vA b = ( b ;~bz).

c

a,

1.

= (2 ; -3) ;

= (3 ; 2). Hay tfnh

@ c ( ; , i).

Vi d". Cho

{GI5 f h o tam gidc ABC b i k A(1 ; I), B(2 ; 3), C(5 ; -1 ). Hay tinh d8 dAi cAc canh AB, BC, CA vA suy ra tam gi6c ABC vu8ng tai A.

= (3 ; -1). Tinh g6c MON.

= (-2 ; - I ) ,

- -- @.ON -6+1 & - -l o ~ ~ . l ~ l = & . 2f i Vey ( O M , ON) = 135'. cos MON = cos(OM, ON) =

'

.

c) Khozing c6ch giaa hai di6m

Cho hai di6m A(xA ; yA) , B(xB ; yB). = (xB - XA ; y~ - yA), do 66 t a tinh d ~ u ckhoHng Ta c6 c6ch 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 :

MN=

I E I = , / ~ = ~ .

d) Bbi todn. Cho tam giAc ABC bi6t A(2 ; 3). B(8 ; 6& C(2 + 4 4 5 ;7). Cho tam giac ABC c6 BC = a, CA = b, A6 = c. Chon he tryc Oxy sao cho g6c 0 t r h g v6i C, di&m B nam tren phdn d ~ a n gc ~ i a truc Ox cbn di6m A c6 tung dd duang. Chdng minh rang :

+ 3) vA

-

1) Tin h do dcii cAc canh AB, AC v2 so"do clia g6c BAC . 2) Tinh dien tich tam giAc ABC.

--

v8 cos BAC = cos(AB,AC) =

Z.A6

246+24& 12.8

Vi3y BAC = 30'.

2) Goi BH 18 d&ng cao c3a tam giAc ABC, (h.2.15). Ta c6 CI

BH = AB sin BAC

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

8. Cac dang bai tap cd b6n Dqng 1. n n h tich v8 hadng cfia hai vecta Phrlmg phap. - DDng dinh nghia tich v8 hudng va cAc tinh chgt c3a tich vd h ~ d n gc3a 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$n l ~ q 18 t hinh chi& c3a A, B l$n giA cBa

CD .

C h m g II. ~ C vo H H

50

~ C~>A G HA1 VECT(I...

Vi du 1. Tam gidc ABC c6 AC = 9cm, CB = 5cm, C = 90". Tinh :

--

a) AB.AC ; -4

b) BA.BC

a) Cach I .

. B

- . , I Ac~ cosA

-_.

AB.AC = IABI

AC AB

= AB.AC. =A C ~ = g2 = 81 (h.2.16)

C

A

Cdch 2. Ta c6 C la hinh chi& cba B 1Cn giA cfia ---.

KC.Uo d6

-.-

AB.AC = AC.AC = A C = ~ 81.

--

. .

b) BA.BC = BC.BC = B C vi ~ 3 18 hinh chi& cba gih cba BC. Viiy BA.BC = 52 = 25.

--

BA

--

tr6n

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

BC = (AC - A B ) ~= AC

Ta suy ra cosA =

AB.AC

IEI.I zI

--

-2

+ AB - 2AC.AB.

-

20 1 --- - - nCn A = 60". - 5.8 2

b) TUcfng tg nhu cAu a), ta c6 :

--

CA.CB =

1 2

1 2

- ( C A ~+ CB2 - AB2) = - (82 + 72 - s2)= 44.

1

Ej2. T~CHVO H

51

~ CUAG HA1 VECTU

C

V i du 3. Cho tam gi6c ABC vu8ng 6 A,

b,

= 60' vA canh huydn c6 g6c BC = 6cm. Tinh tich v6 h ~ b n g cfia c6c c$p vecta sau d i y :

--

a) CA.CB; b)

AB.BC.

h

6o0A

B

A

GIAI Tam gilc ABC 18 tam gilc vu6ng c6

Hin

= 30" nen

1 AB = - BC = 3cm 2

-4 = 436-9 f i 3& (h.2.17) -a) Ta c6 CA.CB = CA.CB.cos 30" = 3 4 3 . 43 6 . = 27. AC =

=

=

2

V i dl? 4. Cho hinh vu8ng ABCD canh a vh c6 t l m 0.'Gpi M mot digm tujl 9 tren canh BC. HZy tinh cAc tich vB hddng :

la

--

a) MA.AB ; b)

E.G.

a) Diing cang thilc hhih chi& t a c6

--M.4.A.B = BA.AB = a.a.cos 180"

lo

b) Goi H la trung didm cGa canh AB. Ta c6 OH I AB. Do 66 :

Hinh 2.18

Chlldng II. flCH

52

HU%GC ~ HA1 A VECTC7..2

Dang 2. C h h g minh cAc dang that v6 vectd c6 lien quan d6n tich v6 hudng

Phuung phdp.

- Sfi dung tinh chgt ph5n ph6i ccria tich v6 h ~ d n gd6i vdi ph6p tong cfic vectu.

- Dhng quy t$c ba di6m d6i

vdi ph6p ceng hoec tril vectu, vi du n h d6i ~ vdi ba di6m A, B, C ba"t kl,ta luBn c6 :

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

. -

.

- - t

. . -

-.A

DA.BC = DA.(DC- DB) = DA.DC - DA.DB

- -- -DB.CA DB.(DA - DC) DB.DA - DB.DC --DC.AB DC.(DB - DA) DC.DB - DC.DA -.-

(1)

--.

=

=

_.

=

_.A

(2)

-_.

=

(3)

V i dl! 2. G Q 0 ~ Ih trung digm clja doan thiing AB v3 M I3 mot di6m tujl ChUng minh r3ng :

v.

--.

MA.MB =

OM^

-O A ~ =

Vi 0 1h trung die"rn cira doan AB n6n

OM^ - O B ~ .

=

-- - - - - - - (OB + OM).(OB- OM)

-s.

Ta c6 : MA.MB = (OA - OM).(OB- OM) =

Vi du 3. Cho hinh ch2 nhat ABCD vh M Ih mot di6m tuj/ ChQng minh r3ng :

MA^ + M C ~= M B ~+ M D ;~ --b) MA.MC = MB.MD . a)

v.

GIA a) Goi 0 18 giao 6ie"m ciia hai d&ng chBo AC va BD cSa hinh ch3 nhet. Ta c6 :

+ M C =~

--

-

(MO + O A ) ~+ (MO + oc12

-= 2 ~ 0 ~ + 0 ~ ~ + 0 CO ~A +, O (C v =i G ) -_.

= 2M02 + OA2 + OC2 + 2 MO.(OA + OC) .

(1)

M B ~ + M D ~( M = ~+SI)~+(EG+OD)~

-- = 2 ~ 0 ~ + 0 ~ ~ + OO ~B +~ O, D( =v ~i )(2) = 2M02 + O B + ~ OD2 + 2 MO.(OB + OD). .

Vi 0 la t6m c6a hinh ch3 nhet n6n ta c6 OA=OB=OC=OD. So sdnh (1)vh (2) ta suy ra :

MA+Md=2Md MB+MD=~=. -- Do 66 (MA + M C ) ~ (MB + O D ) ~ .

=

I

Hinh 2.79

VS du 4. Cho tam gi6c ABC vdi ba trung tuye"n Ib AD, BE, CF. ChQngminh rang :

Vi AD, BE, CF 121 c6c d ~ a n gtrung tuygn n6n t a c6 :

=

-- - --

- [BC.(AB+ AC) + CA.(BA + BC) + AB.(CA + =)I 2

Dang 3. Chimg minh sg vu6ng g6c cOa hai vectd P h m g phap. Sd dung tinh chgt c6a tich v6 hrldng :

iI;d

i.6

= 0.

V i du 1. Chang minh trong met tam giic ba d&ng cao d6ng quy.

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

-HA.BC HA.(HC - HB) .

. .

--

= HA.HC - HA.HB

-- - -HB.CA HB.(HA HC) HB.HA - HB.HC -HC.AB HC.(HB - HA) HC.HB - HC.HA =

--.

=

-

=

---.

=

(1)

-.-

-

.

(2)

. -

(3)

=

--

--

Vi HB I CA n6n HB.CA= 0 vh

--

HC I AB n6n HC.AB = 0. Do d6 tir (4)

Bd

ta suy ra HA.BC = 0 hay Hd I , nghia 18 ba d&ng cao ciia tam giac ABC dbng quy tai H.

Cp 1 I I

Vi dl! 2. Tam gi6c ABC czn c6 AB = AC. Goi H I2 trung di6m cira canh d i y BC vA D la hinh chi& clja H tr6n canh AC. ChQng minh r i n g du'dng trung tuye"n AM clja tam gi6c AHD vu6ng g6c vdi du'ang thang BD.

--

Ta cdn c h h g minh AM.BD = 0 (h.2.21).

Hinh 2.21

Vi M 1h trung digm c6a H D n6n ta c6 : A

_

.

-

2AM=AH+AD A

vA

_

.

_

.

BD=BH+HD

-- - - - -

Do d6 2 AM.BD = (AH + AD).(BH + HD)

--+ (AH -+ HD).HC

= AH.HD

-- -

- ---

= HD.(AH + HC) (vi HC = BH, AH.HC = 0).

Vey AM Vtiang g6c vdi BD. V i dl! 3. TO gi6c ABCD c6 hai d&ng ch4o AC va BD c$t nhau tai 0. Goi H, K I$n lddt I2 truc tzm clja c6c tam gi6c ABO, CDO v2 goi I, J the0 thO tu la trung di6m clja AD VA BC. ChOng minh rang H K vu8ng g6c vdi I).

Chlrmg 11.

56

T~CH H U ~ J GCL~AHAI VECTU...

Hinh 2.22

--- .- -- -Ta c6 2 HK.IJ = HK.(AC + DB) HK.AC + HK.DB Ta c6n c h b g minh HK.IJ = 0.

=

-- - - -

-*

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 sao cho dlrErng cheo AC vu8ng g6c vdi ddirng ch6o BD.

(H.2.23) Tac6:

ACIE

--

oAC.BD=O

-- -

a AC.(AD - AB) = 0

--

--.

csAC.AD - AC.AB = 0

--

Hinh 2.23 -2

Ta c6 AC.AD = AD --.

(1)

= h2 (cBng thGc hinh chigu)

--

AC.AB = DC.AB = ab (cdng thGc hinh chi&)

Do 66

-- --

I % o AC.AD - AC.AB = 0

Dung 4. BiGu thirc toa di) clia tich va huung v9 cAc irng dung : tinh 66 dhi ciia met vecta, tinh khoang cBch giaa hai die"m, tinh g6c giaa hai vecta.

- Cho vecta G - Cho vecta

= (bl ; b2). Khi 66 :

= (al ; a2) vii

-.

u = (ul ; u2). Khi 66

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

- xA)

- G6c

gi3a hai vectu the0 c6ng that :

+ (yg - y A )2 . +

= (al ; a2) va b = (bl ; b2) d ~ q ctinh

Vi du 1. Trong mgt ph%ngOxy cho ba di6m A(3 ; 5), B(-5 ; 1), C(0 ; -4).

-

a) Tinh do dai cdc canh AB v3 AC cfia tam gi5c ABC. b) Tinh g6c BAC .

Vay

-

BAC = 45".

V i du 2. Cho tam g i i c ABC bigt A(-3 ; 61, B(l ; -2), C(6 ; 3). 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 :

b) Goi H(x ; y) 18 trgc t h tam 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 sao 1cho AF =-AD. XAc dinh vi tri clja di6m M tren d&ng 3 t h i n g BC sao cho EFM = 90'.

Chon h$ toa do Oxy sao cho dlnh D trirng v8i g6c 0, c8c di6m C v8 A 16n l ~ q nhm t tren Ox va Oy (h.2.24). Ta c6 D = (0 ; O), C(a; O), A = (O ; a ) , B = (a ; a). Theo gid thi& t a tim d w c toa do A/& B

'I

GiH sB M = (a ; y), ta c&n tim tung do y &a di6m M.

-

Tac6 F E =

(;,;I

-.-

--

E F I FM e FE.FM = 0

-

11Ta suy ra vecta BM = -BC 6

.

Hinh 2.24

59

52. T~CHVO H m G CUA HAi VECTU

V i du 4. X6c dinh g6c gigs c6c cap vectd sau d2y :

=(O;-5);

a)

;

c=(&;l)

- - -

1-1 1-1

a m=p=2.Ta

c) Ta nh$n thgy vectu p = n - m = ( h8y tinh g6c giira

vA

i.

Hinh 2.25

1

1 ' = 2 n6n vecta - - n =m +p +

=

d

-

-

.

4

tao vdi m vA p cPc g6c -

w

e

bhng nhau nghia 18 ( m , n ) = 15' (vectu n = m + p c6 do dAi bhng d ~ l m gch6o 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

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

~ C ~G ,A HA1 VECTCJ...

Ch~rongII. ~ C H H

60

11. Cho tam gidc ABC vudng tai A c6 g6c

6

= 60'. Hay tinh cdc

12. Tam gific d&u ABC c6 canh bang a v2I c6 dllbng cao AH. Hay tinh cbc tich v6 hubng sau day :

C)

-AC. H A .

-di6m M sao cho M A . M B

13. Cho doan thing AB c6 do d i i 2a v i mat s$

k2. Tim tap hqp cdc

= 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 ngoii tam gidc dZ cho c5c tam gidc vu6ng can dinh A la ABD v& ACE. Goi M la trung di6m clja canh BC. ChQng minh rang A M vudng g6c vdi DE.

--

16. Cho b6n digm phan biet A, B, C, D cirng thuoc mot d u h g

-

thing, bigt rang

AB va

IEI= a , l E l =

b. Hay tinh AB.CD khi

ccng hudng v2I ngugc hudng.

17. Cho tam gific ABC c6 BC = a, CA = b, AB = c. VC v6 phia ngoai tam gidc hai hinh vubng ACEF v21 BCDL. ChQng minh :

18. Cho tam gidc deu ABC c6 canh bang a. Hay tinh cdc tich vd hubng sau :

b)

AB.(2E- AC);

C)

(AB - AC).(AB + AC) . -

.

-

.

A

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

-

.

-

-

c) (A6 - AC).(2AB + A D ) . -.

+

20. Cho hai vecta a = (2 ; -4) 4

+

4

,b

+

= (6 ; -1). Tim vecta x = (u ; v)

4

saocho a . x =6vA b . x = 29. -.

21. Cho hai vectu a = (4 ; -5),

6 = (1 ; 2). Tinh c6c tich v 8 hddng

i.6 vA (;+$).(;-GI. 22. Tim g6c giQa cdc cap vectd sau ddy : 4

4

a) a = (4 ; 3) v i b = (1 ; -7) ; 4

4

b) c = (2 ; 5) va d = (3 ;-7) ;

23. Cho hai digm A(l ; 2) vA B(-3 ; 1). Tim digm M(x ; y) sao cho MA I AB vA MA = AB. 24. Tim toa do true tam H(x ; y) clia tam giic ABC bigt cdc digm A(2 ; 11, B(-1 ;3) , C(-2 ; -3). 25. Cho tam gi6c ABC c6 AB = Scm, BC = 7cm, CA = 8cm. Tinh -.-

AB.AC v i tinh g6c A. - C I A

26. 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 :

b) cos 2A

+ cos 28 + cos 2C 2 --.3

2

TAM

I

Chirng ta bi6t rhng mat tam gific d ~ u choan toan xac dinh ne"u bi6t ba canh, hogc hai canh vh g6c xen giila, hoac mat canh va hai g6c k&.

c6 dubng cao AH = h, BC=a,CA=b,AC=c.

Vdi ba truhng hap n6u tren so" do cac canh va cAc g6c cbn lai cria tam giac hoan tohn xac dinh. Nhu v$y giaa cfic ye"u to" clia mat tam giac c6 mat m6i lien he xac dinh nao 66. Nwhi t a goi 66 la cac hg thQc Zupzg trong tam giac. Trong ph6n nay chling ta sG nghien c h nhilng he thI3c d6 va cac & - ~dung g cGa chung.

A

Goi BH = c', CH = b'.

1. DInh li sin

Hay di&n vao c6c 6 trdng sau d i y de" dwc c i c h@ thirc dljng :

a) Cho tam giac ABC vuSng tai A c6 BC = a, CA = b, AB = c A (h.2.26). Ta dS bigt : b sinB=-*a=a

b sin B

a b Do66 a=-sinB sinC = cosB =

-

a

sin L -

c sinC

n

D

'

Vi A = 90"n6n sin A = 1va t a c6 the" vie"t he thI3c (1)nhu sau

a

a -=-=-

c sinC

b sinB

sinA

C

(1) :

'

Hun n3a vi BC = a = 2R 18 duhng kinh duhng trbn ngoai tie"p tam giAc vu6ng ABC n6n :

--a sinA

Ban hag thll xern :

b sinB

- --

-

--C - 2R sinC

He thtlc nay d q c goi 18 dinh li sin trong tam giBc hay d ~ u c goi tht 18 djnh li sin. =

-2I absinC

= di@ntich AABC. Chia cAc so" nay cho

-1 abc ta c6 : 2

a ' b c -=-=sinA sin0 sinC ' Ta d3y hay suy ra dinh l i sin.

b) Dinh li. Vdi tam gi5c ABC ba"t kki, t i so" @Cia m$t canh vA j sin c i a g6c dbi dien v6i canh d6 lu6n lu6n bang d ~ a n gkinh c i a d ~ i r n gtrbn ngoai tigp tam gihc.

p

CU I

a -=-=-

b.

c

sinA

sinB

sinC

= 2R.

5 3 . CGC HE THUC LLJQNG TRONG TAM GIAC VA GIAI TAM GIAC

63

CH~NG MINH myc a) dinh dii d ~ d cc h h g minh trong truang hdp g6c BAC= 90". Ta c8n c h ~ minh g dinh trong tnlang hup g6c BAC kh6c 90". N6u g6c BAC nhon ta d&ng kinh BD ciia duang trbn

d

li

li

CI

vG ngoai ti6p tam giAc. Khi 66 vi tam giAc BCD vu6ng tai C n6n

ta c6 : BC = BD sinD hay a = 2R sinD. (h.2.27a) CC

-

A

Ta c6 BAC = BDC v i 66 1h hai g6c nai tisp c h g c h h cung BC . Do 66 : a = 2R sinA hay

--a - 2R. sin A

CI

Ne"u g6c BAC th ta cfing vE d u h g kinh BD cGa d d h g trbn ngoai tie"p tam gidc ABC. TQ gidc ABDC nai tigp d d h g trbn d 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.

Vay a = 2R sinA hay

Cic dhng thQc

a = 2R. sln A

b -sin B

2R V B

--c - 2R duqc c h h g minh sin C

He

quci. V6i moi tam giAc ABC nai tigp duang trbn bAn kinh R, t a c6 :

a = 2R sinA , b = 2R sinB , c = 2R sinC. c) V i du. TCf hai vi tri A, B cAch nhau 25m v5 cirng thuoc mot ba sBng hay do khoang cdch til vi tri A v i vi tri B tdi vi tri C d

-

b$n kia dbng siing, bigt rsng BAC = 110" v i ABC= 20" (h.2.28).

Chdung II. T/CH

64

vo H

~ C~IA GHA1 VECTIl...

B

A Hinh 2.?*

-

Ta c6 ACB = 180" - (110" + 20") = 50" Theo djnh li sin ta c6 : sin 110' - sin 20' - sin 50' a b 25

Vey a =

25. sin 110' sin 50'

b=

25. sin 20' sin 50'

-

30,7 (m)

z 11,2

(m).

a) BAi tohn. Cho tam gihc ABC bi6t AB = c, AC = b vA bigt g6c A. H5y tinh canh BC = a (h.2.29).

-2

-

-

-2

BC = (AC - A B ) ~= AC

Hinh 2.29

-2

+ AB

--

- 2AC.AB

.,

53. CAC HE THUC L W G TRONG T A M GIAC VA

~ l d TAM l GIAC

65

Tif ke"t q& cfia bhi todn tren ta suy ra dinh li sau diiy : m

h) Dinh li. Trong mat tam giic ba"t kki, binh p h ~ ~ mat n g canh ~j biing tbng binh phdung hai canh cbn lai trllr hai Idn tich hai canh kia nh2n vdi c6sin cSa g6c giila hai canh d6. 1 7

jI

1

Trong moi tam gidc ABC, vdi BC = a, CA = b;AB = c ta lu6n c6 :

L

f [

1-

I

I

a2 = b2 + c2 - 2bc COSA b2 = a2 + c2 -2ac cosB c2 = a2 + b2 - 2ab cosC.

AC = 5. Hay tinh canh BC. a) khi g6c

i= 90" ;

b) khi g6c

= 60".

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

cosB =

a2 + c2 - b 2

theo gi6 tri clja he thOc b2 + c2 - a2 IA mot s6 duong, hoac sd im, hoac bang 0, hay suy ra tinh ch5t cca g6c A (tO, nhon, vudng). C)

Cho 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, C clia tam gibc. Khi d6 ta c6 :

to5

A

m a2 =

Cho tam gidc ABC wbng tai A, c6 BC = a, CA = b, A6 = c.

I Hdy dua "20 cbng t h O 2 ma =

Tinh do dai duirng trung tuye"n ciia tam gidc

2(b2 + cZ)- a2 4

2

I

m,, = 2

mC =

2(b2 + c2)- a2 4

2(a2+c2)-b2 4

2(a2+b2)-c2 4

d6 tinh da d i i rn,.

B

M

C

Hinh 2.30

That v@ygoi M la trung digm ciia canh BC. c6sin vao tam gi6c AMB ta c6 (h.2.30) :

Ap dung dinh

2

a4 = cac cos B m2a = ~ ~ + ( ~ ) ~ - 2 c . ~ c o2 +-sB

Vi cos B = 2

ma =

2

a2 + c2 - b 2 ni2n t a c6 : 2ac a 4

+--

ac. a 2 + c 2 - b 2 - 2(b2+c2)-a2 2ac 4

li

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

b) vi du. Cho tam giic ABC c6 a = BC = 5cm, c = AB = 8cm vA g6c 6 = 77'. Tinh canh AC, c6c g6c A, C cda tam gi6c vA d&ng trung tuye"n mb = BM cda tam gi6c d6 (h.2.31). I A

GI~I Theo dinh li casin.'ta c6 : A C =~ b2 = a2 + c2 - 2ac cosB = 52 +

s2-2.5.8.cos77O

M

= 89 - 80 cos 77'.

-

AC= ~ b2

V$y AC

71.

fi =

Mu6n tinh g6c sinA a

-=-

-

180'

C

8,4 (cm).

2

Hinh 2.31

t a dDng djnh li sin :

sin B a. sin B =,sinA =b b

Do 66 sinA =

e

/A

B

5. sin 77'

fi

-

0,5782

- (77' + 35'19')

-

67'41 ' .

dung cbng thirc :

g

@. Tam gi6c ABC c6 di@n tlch S- N& k6o dhi cqnh AB mot doan AB' = 2AB va kc50 dhi canh AC mot doan AC' = 3AC. G Q S'~ 13 di@ntlch tam gidc AB'C'. S' Hay x6c dinh t1 s6 - .

5

3. Cdng thljlc tinh dien tich tam giac Cho tam gidc ABC c6 BC = a, CA = b, AB = c. Goi R vh r l&n l%t lh ban kinh d&ng trbn ngoai tie"p, nai tizp tam gidc va a+b+c 1B nfia chu vi tam giAc. P= 2 Ta c6 cAc cang thirc tinh di$n tich S clia tam giac ABC sau day : a) Cac cbng thilc :

53.CAC HE THUCLWNG TRONG TAM GIAC VA

~ i d TAM i GIAC

67

abc 2)S= 4R

Ta chlihg minh cdc cdng th3c trCn : A

f

~

Tim die,n tich clia mai tam

1 1)Tam giBc ABC c6 di$n tich S = -ah, v6i ha = AH. Vdi g6c 2 nhon, th hay vudng ta d&uc6 :

AH = AC sinC = b sinC

(h.2.32)

1 Do 66 S = - ab sinC. 2

=

I

l

Hinh 2.32

b)

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

10

2) Theo djnh li sin ta c6 -- 2R hay sinC = sin C 2R '

c)

5

!

I t

C

C

1 1 c abc Ta c6 S = -ab sinC = - a b = 2 2 2R 4R '

3) Gpi I 1P tiim d~lrngtrbn niji tiirp tam giBe ABC. Ki hi$" SaBCla diijn tich tam giBc ABC. BBn kinh r c6a dlllrng trbn niji tigp tam gidc la d~lrngcao xu& phAt til 63nh I cba cac tam giBc IBC, ICA, IAB. Ta c6 (h.2.33) :

Hinh 2.33

Chlrung II. *H

68

Theo dinh li cosin ta c6 cosA = 1 + cosA =

2

(b+c) - a 2bc

1 Vay S2 = -(a 16

+ b + C)

2

@ H U I ~ GC~IA HA1 VECTII. ..

b 2 + c2 - a 2bc

2

. Do 66

:

- (b+c-a)(b+c+a) 2bc

(a + b - C) (b

+ c - a) (C+ a - b).

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) Vi du. Cho tam gi6c ABC c6 AC = 7cm,.AB = 5cm. Bi6t 3 cosA = -, hZy tinh canh BC, dien tich S cGa tam gist, 5. duting cao ha v i b6n kinh R cira d ~ t i n gtr6n ngoa i ti6p tam gidc.

4zii I

ta gpi tam giAc c6 do d i i ba canh li ba so" nguyPn liPn tigp v i c6 dien tich bang mot sd nguyCn li

- Theo dinh li cdsin ta c6 (h.2.34) :

I

3 B C ~= a2 = b2 + c2 - 2bc cosA = 7% 5' - 2.7.5. - = 32. 5

tam g i i c He-r6ng. Cdc tam gidc d6 c6 do da'i c6c canh nhu sau :

Hinh 2.34

4 sinA = -, vi sinA > 0. 5

3

a - Theo dinh li sin ta c6 -- 2R n6n sin A

4. ~ i 6tam i giac v a irng dung a ) GiHi tam giac Gi6i tam giac la tinh cac canh va cac goc ciia tam giac dga tr6n mat so" y6u to" cho tntac. Mu& gihi tam giac, ta c6n tim hi6u giii thie"t vh k6't luan ciia bhi toan d6 lga chon cac he th13c h u n g thich hap (49dduc n6u trong dinh li sin, djnh li c6sin va c6c c6ng thoc tinh di$n tich tam giac.

Vi

dl!

1. Cho tam gi5c ABC bigt canh AC = b = 47cm vA cdc

g6c

Ta c6

= 48',

= 180'

= 57'. Hay tinh

- (48" + 57')

= 75'.

Theo djnh li sin ta c6 : --a - b (h.2.35) sinA sinB j

a=-- bsinA - 47.sin4a0 sinB ,qin7s0

--C -- b sinC sinB

3

i, a va c.

_

36(cm)

A

b=47cm

Hinh 2.35

c=-- b sin C -4 7 . sin 57' = 41. sinB sin750

V 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).

C

Chcrdng II. T~CHVQ H U ~ CGA G HA1 VECTU.. .

70

Hinh 2.36

Theo djnh li casin t c2 = a2 + b2 - 2ak

--- -

c2 = 4g2 + 262 - 2.49.26.cos47' c2 = 2401 V$y c =

V$y

x

+ 676 - 1.737,73 =

Jm = 36,6 (crn).

1.339,27.

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 dien tich S cda tam gidc v i cdc bAn kinh R, r c i a cAc dddng trbn ngoai tie"p, nQi tigp tam gidc.

x,

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

Hinh 2.37

x

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

Goi S 1h dien tich tam.giac ABC ta c6

53. CAC HE THUCLUONG TRONG TAM GIAC VA GIAI TAM GIAC

71

abc abc Ap dung ciing thirc S = ta c6 R = -= 4s

4R

Theo ciing thirc S = pr ta c6 r = nCn r =

S

-. Vi P

p =

13,64 (cm). 24+ 1 3 ~ 1 5 = 26 2

-- 3,3 (cm). 8575

26

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

Trong truang hup cu the" nQy, viec tinh di$n tich bhng cang thirc HC-r6ng cho ta ke"t qu& chinh x8c hun dung cang thirc 1 S = - bc sinA. 2

b) Ong dung vho viec do dac trong thgc ti$ B i i t o i n 1. Mu6n do chi& cao CD = h cda rnqr chi thdp m i ta kh6ng the" de"n d ~ w ctdm C cda chdn thdp. Trong mat phAng dQng chOa chi& cao CD cda thdp ta chon hai die"m A, B tr$n mat da"t sao cho ba dikm A, B, C thsng hhng. Gi5 st3 ta CI

do d w c khodng cdch AB = 24cm vA cdc g6c CAD = 63" , ( C

CBD = 48". Hay tinh chi& cao h clja th5p (h.2.38).

&/ Hinh 2.38

GIAI

Ap dung dinh li sin vao tam gihc ABD ta c6 : AD -- AB = A D = AB.sin B -sinB

Tac6:

I

sin D

sin D

X=E+E=E=X-g= 6 3 " - 4 8 " = h = CD = AD sinA = 68,91.0,891

z

15". 61,4 (m).

Do 66 AD =

24. s i n 48'

z

sin 15'

24.0,7431

= 68,91 (m).

0,2588

BPi to611 2. KhoAng c6ch tCf A de"n B kh6ng the" do trgc tigp d w c vi phai qua mat c6i h8 n u & ~ NgUtfi . ta chon mot di6m C tao nen tam gihc ABC c6 BC = loom, AC = 80m v& g6c CI

ACB = 48'. Tinh khoAng c i c h 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 a = 50'. Tinh hup lgc

vdi

tao v6i nhau mot g6c

v i dnh gbc tao bdi

6 bi6t r5ng ? = 1OON, a = 200N.

Ta vi5 Pd =

-P,OQ -

R

vdi p v ~

GIAI

=

- - Q vh OR O P + O Q . Khi 66 POQ = 50". =

X6t hinh binh hAnh OPRQ (h.2.40), ta c6

-

OPR = 180" - 50" = 130".

A~ dung dinh li c8sin d6i vdi tam gihc OPR ta c6

:

Hinh 2.40

53. CAC HE THOC LUQNG TRUNG TAM GIAC VA

-

cos POR =

~ l d TAM l GIAC

'i3

+ 275i2 - Y ^ ^ ~ - loo22.100.275

0p2 + 0 R 2 - P R ~ 2.OP.OR

7

Ta suy ra POR

2:

-

- 34'

34" vA QOR = 50'

-

16" .

5.Cac dung bai tap cd b8n Dqng 1. Tinh mat s 6 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 h u n g thich hqp d6i vdi tam gi6c de" tinh mat 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"" Vi 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 sin ta c6 : b --=sin A sinB

--a

c

a sinB - 34.sin 45' D o d 6 : b = -sin A ,in71O c=

a sin C - 34. sin 64' -sinA

5

sinC

sin71"

/k

B 2:

25,427

2:

32,32

a = 34

64.)\

C

Hinh 2.41

Chri 9. Ta c6 the" tinh canh c the0 cdng thilc : . c =

b sin C 25,427. sin 64' --

sin B

sin 45'

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

= 48'. Tinh

Chtfmg II. T~CHvo H U T ~ G C ~ HA1 A VECT(7.. .

74

Vgy c =-/,

= 37,714.

Hinh 2.42

x = 99'51'. Tir d6 ta tinh d q c % = 180' - ( E + x) Dhng m6y tinh b6 tlii t a tinh duuc

I

1

f j = 180' - (48'

+ 99'51') = 42'9'.

V i du 3. Cho tam gi5c ABC c6 a = 48, b = 26, c = 30. Tinh C ~ g6c C A, B, C.

Theo he qud cSa dinh li cSsin ta c6 :

Ta tinh duuc

x

i 1 1

A

F

r

117'49'.

Theo dinh li sin ta c6 Do d6 sinB =

i

Hinh 2.43

a -sinA

b sinB

'

b sin A 26.sin 117'49 ' = 0,4791. a

48

Tam giac ABC c6 canh AC ngdn nhgt n6n g6c B nhon. Ta tinh duuc

= 28'37'.

Dqng 2. C h k g minh c6c h e that v6 m6i quan he giGn c;ic y6u t 6 cua met tam giic.

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

h

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 t u m g d ~ u n gvdi mat h$ thGc dd bigt la d6ng. Khi c h h g minh c6n khai thac cac gih thigt vh kgt luan de" tim duuc cAc he thGc thich hap 1Bm trung gian 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 dikm tujl 9. ChlSIng minh r i n g : a)

MA^ + MB2 + M c 2= G A ~+ GB' + GC' + ~

G ;M ~

9 trong d6 m a , mb , m, Idn Iddt 2 I& do d i i c6c trung tuye"n Ong vdi cAc canh a, b, c cGa tam gi6c v i R I i b i n kinh dddng tr6n ngoai tigp tam gi6c ABC.

b) ma + mb

+ m,

2 -R,

-- - --= GA2+ G B ~+ G c 2 + 3 ~ ~ ~ ( 1 )G( Av +i G B + G C = ~ ) . = GA2 + G B ~ + GC2 + ~ G -M2 GM.(GA ~ + GB + GC)

b) Theo h$ th3c (1)ta lu6n lu6n c6

Met khAc ma =

3 3 GA, mb = - GB, m, 2

2

3

= - GC nen : 2

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

hay

9 m,+mb+mc I-R. 2

V i du 2. Cho tam gi6c ABC vudng tai A. Goi ma , mb , m, I$n l ~ la tdo dAi c6c d~ivngtrung tuye"n xu& p h i t tCt cdc dlnh A, B, C cfia tam gi6c. ChUng minh r i n g :

ChUmg II. T ~ H H

76

I

~ C ~G HA1 A VECTO.. .

t

GI~I Theo c6ng that tinh &jdhi 6dmg trung tuye"n ciia tam giac ta cb :

-

i

Vi tam gi6c vu8ng tai A n6n t a c6 a2 = b2 + c2 va ma = -1a 2

V i dl! 3. Cho S

I

IA dien tich cua tam giic ABC. ChUng minh :

1 Ta bigt r i n g S = - AB.AC sinA 2

1

-2

-2

--

= - [AB .AC - (AB.AC)~ I 4

V i dr! 4. Cho tam gi6c ABC c6 BC = a, CA = b, AB = c vA r bAn kinh du'irng trbn nei tie"p. ChUng minh r%ng:

IA 1

tanA c 2 + a 2 - b 2 , a) -- 2 2 2 ' tanB c + b -a B C asin-sin2 2 b) r = A . COS 2

a) Theo djnh li sin va chsin t a c6 :

-

sin A 2R tanA = ---a COSA b 2 + c 2 - a 2 2

2bc ~ ' ~

~

+

~

~

-

~

1

53.CAC HE T H l k LCfONG TFiONG TAM GlAC Vi GlAl TAM GIAC --

77

-

-

-

-

b sin B 2ac 2R - -b tanB = -COSB a 2 + c 2 - b 2 2 R ' a 2 + C 2 - b 2 tanA a 2 + c 2 - b 2 Do 66 ta suy ra -t a n B b 2 + c2 - a 2

'

b) Goi I 1h tAm d ~ a n gtrbn nai tigp tam giAc ABC. Cdc canh BC, 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, t a c6 -- sin - sinBIC 2

--

~~

-

sin

+ 2

- BC -B+C A sin - cos 2 2

-

(1)

Hinh 2.44

MGt k h i c trong tam giAc vuang IMB t a c6 :

IB =

. B' L

sin 2

Tir (1)vh (2) ta c6 :

1

B

--

C -

sin -.sin 2 2 B C a. sin -sin 2 2 Do d6 : r = A cos 0

i l

COS

A ' 2

:

Chllang II. T/CH

78

vo HU%GCUA HA1 VECTII.. .

V i du 5 . Cho tam giAc ABC c6 C = 6. H%ytinh :

A

= 60°, canh b = 12,

a) Canh a v2 g6c B clja tam gidc ABC ;

b) Dien tich S clja tam gidc ABC ; C)

B5n kinh R clia dllirng trbn ngoai tigp tam giac ABC ;

d) €38 dAi d ~ d n gcao BH vA dildng trung tuygn A M cGa tam giic. ,'

GIAI

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

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

1 2

fi

= -.12.6.-=

2

18&.

Nh@n xkt. Vi g6c B = 90" n6n ta c6 the" tinh dien tich tam gihc the0 c6ng thBc :

I I

60' A

1 S = -.AB.BC = -.6.- 1 2 & --1 8 f i . 2 2 2

H

b=12

6

Hinh 2.45

abc - 6&.12.6 c ) T ~ c ~ s = * , ~ o ~ -~ R = = 6. 4R 4s 4.18.A

Nh&n xkt. Vi tam gi6c ABC vu8ng h i B nen bhn kinh d ~ i r n g 1 trbn ngoai ti6p tam giac bhng - canh huy6n AC. Ta c6 2

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

Darlg 3. Gi5i tam gi5c v,i v s n d6 do dac. l'h rltrng ph cip

- Tam giac ABC t h ~ l r n gduuc cho dudi ba dang chu

y6u : bi2;t mot canh va hai g6c k6 (g, c, g), bi6t mot g6c va hai canh kg g6c do (c, g, c), va bie't ba canh (c, c, c).

- De" tim cac yku to" cbn lai cua tam giac n w d i ta thuirng dung cac djnh li sin, dinh li &sin, dinh li : "Tdng ba g6c cha mat tam giAc bhng 180"" va dac biet c6 the" dhng chc he thdc l w n g trong 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 tam gi6c ABC d6.

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

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

V i du 2. Ngllili ta mu6n do khoang c5ch tir hai vi tri A vh B c6ch nhau 500m 6 ben nay bd sang tdi vi tri C 6 bi2n kia bd

-

sang. Ho do d l l ~ cg6c CAB = 87", g6c CBA = 62". Hay neu c6ch tinh cAc khoing c6ch AC v21 BC (h.2.46).

Theo dinh li sin :

a ---b C -sinA s i n B sin C

ta c6 :

I'

c h ~ m 11. g T~CHvo H U ~ I G C ~ , A HAI VECTCJ...

80

Hinh 2.46

c. sin B - 500. sin 62" A C = b = -sin C sin 31"

c. sin A B C = a = -sin C

-

-

i

857(m)

500. sin 87' = 969(m). sin 3 1"

V i du 3. Mu6n bie"t do cao clja t h i p cao nhst the" gidi 6 Toronto thuoc Canada (c6 t2n goi I i "Canada's National Tower") ngddi ta ti& h i n h nhd sau :

Goi S I i dlnh thdp v i C I i tam c6a chan th5p. Vi khbng do dduc ti, chdn C clja thip, ngdtfi ta lgy mat digm A n i m tr@ndddng n i m ngang vdi chAn C clja t h i p v i do dddc g6c SAC = 70". Tr@ndclZlng thing CA ngddi ta IA'y mat digm B c6 khoAng cdch AB = 118m vA do dddc g6c SBC = 60". Ti, d6 ngddi ta tinh d w c chi& cao clja thdp (h.2.47). HGy cho bie"t ho d2 tinh chi& cao nay nhd the" nio.

Mu6n tinh do cao h = SC ta cAn bi6t d q c do dhi canh huy&n SA cGa tam gihc SAC vu6ng tai C va c6 g6c = 70". X6t tam giAc SAB t a c6 ASB = 70" - 60" = 10".

S

C-

Ap

dung dinh li sin d6i vdi tam gihc SAB ta c6 : --AB sin 10'

-

SA sin 60" Hinh 2.47

Do 66 SA =

AB.sin 60" sin 10"

-

118.0,86602 = 588,5(m). 0,17365

:

.

Vi sin 70'

=

SA '

t a suy ra SC = SA.sin70°

= 588,5.0,9:

the"gibi lh 553m.

V(iy chi& cao SC cia

27:Cho

tam gidc ABC bigt :

a)

=90°,

b)

= 90°,

6

=58', a=72cm.Tinh

e , b, c ; A

= 48', a = 20cm. Tinh C, a, c.

28. Tinh gbc Idn nhgt c i a tam gidc. ABC vA di@ntich clja tam gi%c d6 trong m8i trcldng hgp sau : a) a = 4 0 c m , b = 13cm, c = 3 7 c m ; b)a=13cm,b=5cm,c=12cm. 29. Cho tam gidc ABC bigt cosA =

3 5

-, b = 5,

c = 7.

Tinh a, SAABC, R VA r.

x,

30. Tinh cdc g6c v i d d n g cao ha, bdn kinh dlllrng trbn ngoai tigp R, noi tiG'p r clja tam gidc ABC biG't :

& , b = 2 , c = JS +I. 31. Goi G la trong t;im clja tam giAc ABC. ChQng minh r?ing.: a=

32. ChQng rninh r3ng trong tam gidc ABC, ta cb :

+ c.cosB ; b) sinA = sinB cosC + sinC cosB. 33. Tam gidc ABC c6 chu vi 2p = a + b + c. ChQng minh rsng a) a = b.cosC

.2 sinA = ,-Jp(p bc

- a)(p- b)(p- c) .

34. Goi m a , mb, m, I8n l w t Ii cdc trung tuye"n Qng vdi cdc canh a, b, c crja tam gidc ABC.

a)Tinhmabie"ta=26,b=18,c=16; b) Tinh c bigt a = 7 , b = 11 , m, = 6. C)

ChQng minh rang 4(m: + m i +m:)

= 3(a2 + b2+ c2).