z = 4 (E)O: (D)O: (B) (C) (A)O: 1.(A)(C)(D)(E) 2.(A)(D)(E) (B) OAOBOCxyzABC ==⇒= ::cos:cos:cos yOCBzOAC == cos,cos

z x

y

D F E

O A

B C

x x

z

z y

y

F

E D

O A

B C

一、

1.(B) 2.(B)

解：如右圖所示，

∠

BOC

= ∠ ∵對同弧的圓心角是圓周角的 2 倍) 2

A

( ⇒ ∠

BOD

= ∠

A

(∵Δ

BOC

是等腰三角形)

∴ =x OBcos∠BOD=OBcosA 同理，

y

=

OC

cos ,

B z

=

OA

cos

C

又

OA

=

OB

=

OC

⇒

x y z

: : =cos

A

: cos

B

: cos

C

故選(B)

二、

1.(A)(C)(D)(E) 2.(A)(D)(E)

解：(A)O: 由投影定理知，c=acosB b+ cosA (B)×:同理，b=acosC+ccosA

(C)×:同理，a=bcosC+ccosB

(D)O:由餘弦定理知c² =a²+b²−2abcosC⇒ =c a²+b²−2abcosC

(E)O:由正弦定理知 sin

sin sin sin

a c C

c a A

=

C

⇒ =

A

3.(A)(B)(C)

解：(A)O: 7 8 9 2 12,

s

= + + = 由海龍公式 Δ

ABC

= 12 5 4 3⋅ ⋅ ⋅ =12 5

(B)O: 9 7 8 21 21 5

4 4 12 5 2 5 10

ABC abc R R

Δ = ⇒ = ⋅ ⋅ = =

⋅

(C)O:Δ

ABC

=

rs

⇒12 5=12

r

∴ =

r

5

(D)×:由右圖及計算結果可得x= −s BC=12 9− =3

又

64 9 2 64 56 81

cos 2 8 3 2 8 7

A

= + −

BD

= + −

⋅ ⋅ ⋅ ⋅

2

73 39 2

7 511 117 394

3 7

394 7

BD BD

BD

⇒ − = ⇒ = − =

∴ =

(E)×:我們也可以得到

z

= 4

^{9 5 14, 16 5 21} : 14 : 21 2 : 3

OA OC

OA OC

= + = = + =

∴ = =

故選(A)(B)(C) 三、

1.(1)7 (2)1

2 (3)39

4 3 (4)49 3

π

2.(1) 4 6 (2) 6 2 3. 6

−12 4.1

4

解：∵底邊的比會是高的倒數比

AB

2,

⇒

BC

= 令

BC

= ⇒2

AB

=4,

BD

= 1

1

cos 4

B BD

∴ =

AB

= 5. 7

10

解： 1

6 sin

ABD

2

AD BAD

Δ = ⋅ ⋅ ∠

1

7 sin

ACD

2

AD DAC

Δ = ⋅ ⋅ ∠

又 3 3 6 sin

5 5 7 sin

ABD BAD

ACD DAC

Δ = ⇒ = ∠

Δ ∠

sin 7

sin 10

BAD DAC

∴ ∠ =

∠ 6.

解：如圖，在 AMBΔ 及ΔMNB中分別應用餘弦定理，得

2 16 80 32 5 cos 96 32 5 cos

MB = + − A= − A

及

2 16 16 32 cos 32 32 cos

MB

= + −

N

= −

N

兩式相減，得 cos

N

= 5 cos

A

− 由三角形的面積公式。有 2.

S

²+

T

² =(8 5 sin )

A

²+(8sin

N

)²

E

C A

B D

4

4 5 T S

M

A B

L

γ γ

β α

D E

F A

B C

L γ

β α

E

F

D A

B C

F E

A C B

2 2

2 2

320(1 cos ) 64(1 cos ) 384 320 cos 64( 5 cos 2)

A N

A A

= − + −

= − − −

1 2

640 cos 256.

5

⎛ A ⎞

= − ⎜ − ⎟ +

⎝ ⎠ (1) 四、

2.

證明：設截線 L 與三條直線

AB BC CA 的夾角為

, ,

α β γ

, , ,如下圖所示。

(1) 由正弦定理得

sin sin

sin( ) sin

sin sin

sin sin AF EA BD FB CE DC

γ α

π α α

β β

β γ

=

= − =

=

sin sin sin 1

sin sin sin AF BD CE AF BD CE FB DC EA EA FB DC

γ α β

α β γ

∴ ⋅ ⋅ = ⋅ ⋅

= ⋅ ⋅ =

即

AF BD CE

1.

FB DC EA

⋅ ⋅ = (2) 由正弦定理得

sin sin

sin( ) sin

sin sin

sin sin AF AE BD FB CE DC

γ α

π α α

β β

β γ

=

= − =

=

sin sin sin 1

sin sin sin AF BD CE AF BD CE FB DC EA AE FB DC

γ α β

α β γ

∴ ⋅ ⋅ = ⋅ ⋅

= ⋅ ⋅ =

即

AF BD CE

1.

FB DC EA

⋅ ⋅ = 3.

證明：在 ABEΔ 及ΔACF中分別應用餘弦定理，得

2 2 2

2 cos

AE

=

AB

+

BE

−

ABBE B

(1) 及

2 2 2

2 cos

AF

=

AC

+

CF

−

ACCF C

(2) 由題設 1

4 ,

BE

=

CF

=

BC

於是 (1) (2)+ 得 ^{2 2 2 2} 1 ² 1

( ) ( cos cos )

8 2

AE

+

AF

=

AB

+

AC

+

BC

−

BC AB B

+

AC C

2 2

2 2 2

2 2 2

1 1

8 2

1 1

8 2

9 4 5

8 8 8 .

AB AC

BC BC BC AB AC

BC BC

BC BC BC

BC BC BC

⎛ ⎞

= + − ⎜ + ⎟

⎝ ⎠

= + −

= − =