1972-06-08 — Page 17

ENCE LIBRARY

育教僑華頁一第張五第 日七廿月四年子壬

9 JUN 1972

WAH KIU YAT PO

{ då CITY HAL

四期星

197271

日八月六年二七九一年一十六國民華中

BD

a(25mA COA)+

ZSMA

•b (25m BGD BY

25mm B

張正邦

VER

+654-BT. EF + FD + DE = a cop +6 008+ccoc

張正邦 •

(7) (a) $41.25in20−(2+√3)&no=2ti

*+0°ses 180o.

25mm2 0-(2+)60=2+13,

2(1~6530)-(2+ E)∞©=2+£3,

2-2020 ~(2+√3)60 =2+√3,

2 €26+(2+B)CO +3 = 0. (2CD8 +√3 ) (c2O+1) = 0

•1 2658+√3 =( $COB+1=0 BP Coo

Goo

.*. 0=1500 X 8=180°

(0°≤0≤180)

b)有一梯,斜放於牆上,該梯與 地面所成之角為人當梯頂從 牆上垂直滑下P呎時,梯脚移 動q呎,而該梯與地面所成之角 *B*P = 8 ct = (X+B).

CE = XR; AC = 4 be

nts ACD: AD=(P+X) cxx.

its BCEP BE = X C&C B

AD = BE =

梯長

'(P+X) C&<<<= X, C&C B,

·EP X = Placa

uta, BCE

P&im

(1)

Cel B-Lock

Sund-Lin B.

·8+Y= BC= X Cot ß

PSin B

COB PCDB

-(2)

(3)

POB

Cind-Sinß

L

Sund-Siß Simß sind Shop

At.A ACDP Y= (P+X)∞tα=

Ma +P.+x) Cotx=

·8+Plot x+x6tα = Flooß

Sind-Siß

(元代入):

+ Plota +

PSimB Simon-Shaß

6630

PCOB

Sind - Smß

PSigß Gou

(t)

=

Sind-Sinß Sind (Sind-Sim B) Sund

PSMK GoB-PCOKS inB-PCord (Sind-Siß)

Sim∞ (Sind - Sinß."

=P..

SMKC GOB-Cox)

Since (Sink-Sink)

25 mm - Sin B.

•

:-Plan (2+#)

P=7 lot = (X+B).

仕死鬥三形ABC中,DEF為三角形 三之垂證明

~ EF a CoA. £t a BCLLE

B

DEF=180°-2B,

-DEF2##2

C(25m C. CoC). 25mc..

==(in Sin2A + to Sun2B+

SB

Su2()

==(2RṢm2A +2RS 2B +2R Sun 2C)

R(Sm2A +Sm2B +Sm20)

故得證

Buh tan A, tan B; tam C

can (A+B+c).

tan (A+B+C)=

tan A+ zan (BTC

1- tan A tan (B+C) tan B+ tan

tand + tamb tanc

1

tamA (1-tan Blanc) + tan B + Jan C

中文中學會考試題預習專欄 數學科 (卅二)

(續上期預習題解答

40.PDE上一點C2切緑平行カ 弦DE.西弦PDBPE之進線相 ** A, B ZA ZL AC: CB - AD BE.

Eko ABIIDE:

ACB為PDE之切線

*LE AC CB=AD • BE 証明

PC, DC, EC.

I. ACB TOPPE (EKO)

2 21-23 (Rash że wg5 Dan)

3. AB DE (24)

"

tama tanB + tanc

1- tamB tanc

+ 22=23 (MAIND

5. 2|=22 (WIR)

6.

24=22 (#B3 2QRĦAX)

(1-tomB tan()

·tan ♬ ( tam B+tan ()

7.

41-44 (0,0**)

tam B + tamB +tame - tam A tan Blanc.. - tame tan- teme lanĤ — fouÀ H.R

(A, B, C = A

Z tam A + tam B+ tan (= tan Atan B tom

A+B+C=180°

2

It can fit tam B + tam - tam A tams tax By tan (A+B+C)=0 [F]#{(a)#**] A+B+C=0° A+B+C=18 A. B. CLA+B+6=0 5. A+B + C = 180°- (C)在直角三角形ABC中,AB海斜邊.

·BC=3AC ABC IK # P.Q

12 BP=PQ = QC. *B

<BAC+2 PAC+LQAC = 180°

A

DOCH AC=CQ=QP=PB=X

Dan <QAC

AC-=1

CB

SA

Tan PAC=

CanLBAC

應用(b)結果,

tam &QAC + tama PAC + Iam <BAC

=(+2+3=6 =(1)(2)(3)= IamLRAC tam LPAC Jamz BAC.

24QAC+¿PAC † ¿BAC **** 180*

九試繪出下列两方程之圖緑..

.4 = Sin 20

√EGO-Sine.

o's e≤ 180°

AP: BP ACBC (= (443-17 2

$1

鄰邊成比例)..

1 AP: BP = AD : BE.

10. AC÷CB = AD · BE (@.@*IË)

BED

XB2 = 2D = DR (@_@K #) ABZ ·AC

CD2: AB = DB CA

56. BAKA, B, X&ABLEMEL

EL,M; XP XM

図PQA,LMA相筴族5T試証

PS=MT

Ek9 BAQP SEJALM

KAB.

XPLOM?

AN

#21. PS=NT:

I XA XB=XP⋅ xsrael-ele

XA XB = XM XT 2014

部之箱).

2 XP-X5=xMXT

3- X E PLOM

4. XP=X. (2 48

已

愛代換】

5. XS = XT (34除等等)

6. XP=XS+SP, (48IK#sti#

XM=XT+TM

7. XS+SP=XT+TM (18x/R)

TM (7-3量減等量二 By Ps=MT.

16. ABC為一角形一直線與BC之延長8. *CA, AB 51 **P. Q. R; Cx $ 13 * PQ HE AB * £KEX, 30 DE

Be

i) BP = BR

氏定理).

BPXC AR

A.E.L

BP x CAX AR-1 ({ ba. 1=MBACCOS AB, ACI BERE,D

Bko RPMXC

પ

P.

CP

C.D

BR

XR

O BP x LA, DK-1

1. XC/ RP (zke)

2. BC

CP

3.

BX (941 A ≥ 216,

分两腰成比例2. (ABE)

BCTCP BX+XR

•=

ср

XR

4 BC+CP = BP, BX+XR=BR.

(445)

5.

第二筆(4代入3) :. (A = x2 (RO)

RA

BEX CA - BE XXX (#1476, (化簡QED.

8 28 x x =

su. ABC内接於一国之通形し

De

1 BE = AD. * BD. CE OS POS

辣躐

D

BAR SABCDE=H.

*. BEAD OF

BD, CE KK POS

1設合於所設條件的點落P,則在

A HED TO ABCE & AB=BC, AD=BG.

* BAD = LEBC #66. • SAS DABD SABCE.

LADB =<BEC 62 EAE.P.DE-

ЯE. LEPPALA =168° ★★ LBPC =

LEPD = 180° - LA = 188" - 60" = 120°. By 1°. 點在以BC為弦120日同側的 巴形網上。

D BLUSAS 34 BCE HER SP

BP. CP.AC. ABID.E

RIED ABY THOYCE 9, 144 BPC = DO", LPBC +2 BCP = 60 13 2 PEC+LPBE

=60AL ABD = LBCE, 8 LA- LEBC÷6 AB = BC ASA DARD & ABCE, ~. FD=B

即ABC上的點,都合於所設條件

點え切線質超B點平能線 2由12知可求的軽酢是以腔

#⭑*D. * CD*: AB2=DB: CA.

Eko ACUBD.

Y=C(8+30°)

0'se≤ 180

並利用該兩個線解方程

DC ABC 2

切線.

*. CD2: A8*

製表如下

DB: CA.

20 30° 45° 60*

Size

1 DC BE ABC 2 to bk (250)

2

O

O 98° 120° | 135*. | 146* 150* 168. Sin 2010-0866 −0.984) −0.8664 -0.7460 O

Send-0840 -09669 −0.9868

180°

3.

ACH BD

-/-apps-02660)

0°

O ļa 3420| 0.6428|0,8660| } 10.8860] |6(0+30)0.8660 07660 06428 05000 0.258

-|

y=Sina

15

LEFD=180°-2C, LFDE=1860-2A.

又圈DEF之半径=ABC之半径之半星

魔用正弦定律EF

SinAFDE=DEF

**95*=2()=K., '

EF = R. Sin(180° -2A)=R Sin 2H

R(25mA∞∞A)=(RSUNA) (PH

= a CoA, (VOHBCP $=2R) 故得證

(UD) = AS DEF = R(SMZA+

SmZB+ Sm2c)‡‡R*##@ 之半径

Don Er- Q COA.

b. Co B

Jš ane-Sme = 25m20 •TIC#

√ Cose-2 Sine == Sinz0, of Crs 30000

-

*. DE DC, BALLO F

FC2= FB.FA CO§4-X:\-H

切之平方於由點子園主張 全部與外部之積)

(eko)

4. C2 = PA (-142445 =§¤§ -12.

照(一直線平行酒形一退

FC FA

則分他两基成比例)

5

CD = 6 (****) (端平分)

6.

FA

CD2=FBA

FB:A8* ( @שHLH)

FP

7. CD2 = 28 (19)

FA

2DBF=2CAP, (ank mutatɔ. 4BDF=LACF.

NOACF..

8.

1. BDF ~D.

同理.FD

· Sin 30"Sind = Sin 20

to

RD

DE = CGD C.

(禾完。轉第四張第三歲)

AC

則此两三角形相似)

ER(相似三角形對應边成比例)

FR

3à. € 12, WA12) 65 2 HE 30 BC

!

66. #CABE *. HC = | cm.

CB=3 City - BB 20 Q. BAB Tami C. BAB 143.obfu≥akan 求此园之半程

AC=1 ami

CB=3 cm!

AB$112.00

10

₤r.

1. AB=AC÷CB. 41¥Kasti**

い

2. AC=7em. <B=}Cm. (Z k*);

3. AB=7cm + 3_cm. = 10 cm.

4. OP==AB

5. OP = 1x10 can. = 5 cm:

6.

OP = 0Q+r (10)

7. 0Q=OP-Y___ (314)

8.00=19-Y) (m. 1 @t£x@]

9. Aot oc=AC

10

It.

13,

·OC=AC-AL

OC=(7-5) cm. = 2 cm.

12. FAQ0C 0Q2= oc2+ r2 (\ÑÑŘ¥)

(5-1)2 = 22+ y2 (@®.04^Q)

r=24N).

14