ページ1:

L
面積:
L12 三角比性質
公式:
1. DABLD 1 AL = 8 80 = 6
- tab. I th
D
正弦
a
C=2R
SinA - SinB = sinc
a: bc = sin A = sin Bisinc
a²= b²+ c²-abc cosA
SABL = be sin A
abc
48
=
rs=5(SA)(-b)(
11: 1 x 6 x 1 a + 1 x 6 x 51(8-5)
=
a
351 +1253 -
a
= 1253.
A
B
2. ABC LABL=60°, BL=3 BA=S
求CABC的分角線段長
角平分線:知角度左十右二全(正)
不知角度共角解(餘)
B
BD = X
3
A D
±5K sin? + 7x sin 30 = 15 sinb."
153
灾了知三角形中線求面積
B
E
A
☆4.知三角形高求面積
PK
A
Jk
X = IT√3
面積:延伸箱和中線立一樣長
△BDE=AOEC (SAS)
△BOD=海龍公式:ABOC
OD = = AE 4x3 = DABL
BD=0C=LF
設ha:厂面積:求出△ABC三邊比
h=b
AP = AL = FL = T: 5 = 6
B.
bk
J
hc=b SABL = №8k. 3k- 3kk = — — bk. 5
K = 4 GABL=
75
4*
Double A.

ページ2:

3
1/24
5.三角形T三邊3.4.x另一個23,4,x+2
T₁
T₁ = T₂ (1) tex.
T2
1 = 1 x 12 x sinα = 1x1 sin B
4
3.
P
x+2
16+9-(X+21~
6.
B
7-
A
"
↳ sinα to sing to (1)
24
cosα= 14+9-x- =
x= -1+256
AC + B0 = 2 (AD + CD").
=
24
A₁ = a+b²-2ab cos
O
大差負
消
+ 80 = a² + b²-2ab cos (12-0)
"AC" + BD = 2 (a² + b²)
√5 (5-9) (5-6) (5-c) = D
SABC tabsine
=
[(zabta+b²-(^) (pab-at-
+c)
=
ab √√l-cos L
=
= 4√(a+b)-cˇ][c²- (ab)]
=
√ (a+b+c)(atb_c)(a+b+c)(-a+b+c
√
2
2
7)
=
8. SSA題型:
A
29b
4 √(zabi`- (a²+b² - c² ju
ABC + a = 1 b = √3 LA=30"
A
3.
S
C
S(S-a)(S-b) (S-C)
C可能是30,90
(可能是1或2
Double A

ページ3:

9. b (b-c cos A) = a (a-ccos B) AI
B
sol: b (a cos() = a(bcos ()k
b=acob+bcosA
6- ) = ( 45-6)
bla
bta-c
Ayd
7Aby
h
A
Ccos Ab a cust
10. A ABC AB = 7 BL = 8 TA = 9.
向外各作正方形得ABNM、ACRS
BCPMS OR NP & M N P Q R S
MNPQRS = 48ABL +7 +8 +9"
L
M
= 4√12·5.4.3 +7 +849
= 194+4855
11. ABCD & AB = 6 BU=4
CD=2LABL=LBLD=120°
B120
JAD
廷伸AB、CD交日
E
120
4
ABECIA
AD = √10²+6" - 120 cosbo = 2√194
=№10+6-120
12.6 ABC AB = 5 BL = 6 AL = 7
LA之外角平分線交能於D
JAD
(2.5
B
b
9
J
Q
角互铺
Sin 相同
7
确定理:
A = AB==BD
725 = b+x=x
X=15
AD = √21+7-1470C
Double A
= 250
D D
B
C
L
5 = 7 = BD = 20 (1)

ページ4:

13. A BLD AL = 8 BD = 10 2
C和BD銳夾角60求面積
B
A
D
14. 平行四邊形ABCD
AB = 6 BL = 10
sol: x 8 x 10 x sinbo
=2053
1
角10 求面積
A
D
b
s.1: 036 = x²+y-2xy cost."
100 = x²+
-2xyco5120"
+100 = x²+1+xy cosb.".
64=2x4 XY=32
2-1
ABCD 4 Lysinbo') = 3√3
10
Double A
O O O O O O