4 cos cos (90
= cos [(90°-a)-(-B))(_b/AD!
=
EF=Sin
(2) sin(a+B)
=
說例2右圖中,□ ABCD 為矩形,
Usen 发布形,F-1, 2 BAE =a,
( ZEAF =B LABE =
90°=∠AEF,則,並點食?
x+ß
(1) Z CEF
AD =
( 90° − a)cos (− B)+sin(90 ° — a)sin ( — ß)
a 30
= sin a cos ß- cos asin ß +
LAFD=0
-0) 200
E (8-0) wiz
CE-sinpcosa, BE = cos sina
BETTE=sinacosp=05
9
• AE = COS
COSBO,
D
F
=
3° △ABE中,AE cos BBAE =a 2B = 90°
01 200
⇒ BE = AE sina = cos sin
a
C
Sing sing cosa
E
a+B/
cosp
α
cospsind
B
行
ODN12
cosasing
【】: \\°_ < AEB +a= 90°= ≤ AÉB + 2 CEF :.< CEF = a
EF LCEF=0 家企业:[意莹]
DJO
LAFD = 2 BAF (=a+B
在△AEF中,∠ AEF = 90°,斜邊 AF = 1,
LEAF =B ⇒ AE = cos ß EF = sin ß
9
9000°78 0p) me! ( )
= ('01-'04) 205 S( )
b=
4° △ CEF中,∠CEF = 90°,斜邊EF=sinB -
美意()
194
⇒ CE = EF cos a= sin ß cos a 0 aiz+01 2000 205 =°08203 A( )
9
5° AADF, 2D = 90° 2 CFD=a+B AF = 1) 200=31 203.2( )
⇒sin(a+ß)=1 CD= BE + CE
⇒sin (a+ß) = sin a cos ß+ cos a sin p
Ab
L₂ E
CIEJ
1+
45°
15
m=
||
OK!!謝謝!!