BESHAR
日元初月十年時門聲護
WAH KIU YAT PO
A
FRAME 會考試題預習
4 ABD
BD $
·5 x 4 cos 56°16:
期温
日八廿月一十年八六九一公年七十五國民申
585
數科 (四) ·喬仲强
第三次蕷智題解答
川謖日篤鋭角試用三角速馼え銭跂定義証明
ain (180°-0)=sino
cos ( 180° – 0) = -co☺
tan(180-0)—-tang.
(3) than to LAOP-180°-0
$1 <POA-034 0237129
【半径=1)ATBS 为0园切线,
PMIX軸 PNLy軸位三角函
數线段定義得。
sin (180°~0) =MP (I)
cos (180 - 8)
tan (180-0)== AT
cot (180°-0) — BS)
sec (180-6) FOT
cac (180 - 8) = 05 (1)
cot (180°-6)=-coto
acc (180~0)————sec 0.
cse (18029) — NCA
B
延長 PN交口於P联OP并延長之支Å要切线
*T 2 BE DIS KOTIM
it o ONP rt AONP (OP== OF ON 為公共场
x rt OBS = st.0085' (8) <5'08=LBOS 08$&tui
LAOT=¿POA (XITLA) LADT=2
te ht SOAT FAC BOAT B rt A OMP ZAL A AM
sin MP. Cos H
tan =
洋各銭続的適正値由上面接証者対金等三角形中
24 MP = MP
07-07
sin (180°
-NP AT =
cot (180-0)=-cute,
-25+16-40x 6. 5 $5 3 — 18, 79
BD=18.79. 434(準至三住有效數字)
~BD=434
(6) DABC, BC=6at, CA=8ot, AB=40t D&BC
L+ - $m CD = = = $ AD=
(解)在△ABC中
·Cas B
CLA ABDY.
·AD-
1*- 2×4 × 4 CAIB.
LLO
AD=140=2/10 = 6:32
(7) ▲ ABC+ AB- 5 at AC= 40
的要心HRA奌的距離
(角談如右園在がABC中
BC= 4+5 - 2x4x5 cms 108°
=16+25+ 4s cos 72
=40x 0.3090.
=5336
80-$337.305 修正弦定律
30.5
| # $ X X #2 5 3 #3
僑
*#$%Ŧ 45%£ / SAS SK3#8%rns #
PT%$#$%$#D%S# $%$#$%$# $%$# $£$#$%$# 5%$#!
(接第六張第二頁)
文中學會考
【題預習
生物科
四
廖百 R
·AD=2/10 A 6.3.2 <BAC=108?
ABC
Wreen plants use oxygen chierly in the process of (1) photosynthesis (2) germination
(3) transpiration (4) respiration (5) Absorption
5.
The laboratory procedure for removing
hlorophyll is to boil e green leaf in
(1) alcohol (2) iodine solution (3) tap water
(b) 14me water (5) caustic soda solution.
The young pient in a seed is called the (1) cotyledon (2) endospern (3) hypocotyl (4) epicotyl (5) embryo.
or the following, which is the monocotyledonous seed?:
(1) Eroad bean seed (2) maize grain (3) French bean seed (4) castor oil sead (5) soy bean seed.
8.
5 sin 108 7.305
Gain 22 $305
NO:
5
0.6990
log sino,-T8136, 1€ 0.4038" lain 70° 7978264
0.67.72 <BHA=
= 40° 38′ ( F £ <H BD 1); 7: 305 0.8636 = 47°22 (E] <D=M 2)
=5838
Shich one of the following seeds carries out epigeal germination".
17.8/.36
(1) Broad bean (2) maize (3) pea seed (4) castor oil seed (5) soy bean 90
<HAB = 180° - Q, =121°22 (10 HAD 3- 11t
10.
alled (1) cotyledon (2) endosperm (3) embryo
10. (4) testa (5) hypocotyli
cos (180°
tan (180° – 0)
€
tan o
nac (180 -8)
caco
(2)(a) & ABC + $1
(頻就B角馬鈍角
a 鋭角
& cos C + C COMB
别証明)
<HBA d2- 2 BHA - 58° 38′ – 40°38′18′′. ☆ HBA & 16 1 1 1 3 1
AH
NO
5
LOG 0.6990
sin 1874900 (+
0.18.96
45° 38′ 7.8.136)
XAH=2·37 of
0:3754
(31) # BAB SLA (4 8)) w
12 A 12 ADI BC 11 4 ABD 4
BD AB cos (180-B) == cos B
cor (180 -8)=
BD AB
-bcos
REA ACD
· C8—CD-8D—
(-(-cco B) = bar+cos
*BA CAMSIA (FB) 14 ADL BC, B
BD = AB cos B = c cos B · DC AC cos c = b cos C
BC=BD+DC=cces B+ & cos C..
BA BALAA AL} a=bcosc+cco¥B.
(B)利用上面所得结果,証明餘弦定律
a = b2+c2 -2bc css A⠀
(正)由上面得
同理
(1)xa.
(3)× C
(5)+(6)-(4)
4-1
a = boos crc COT B
—a cos PTC COSA C=aces B + b Los A a=abcos Craccos B b=ab cos c + fc cos A c2= acco2B+ &C.COLA
b+c-a
28 c. cos A
f+c-2bccos A = 0.
(3) ▲ ABC Ý, 3★ SE
(a) (b+c) cos A++(c+a) cos &+(a+b)ños (= (誰)由上題時得(1)(2)(3)武相加
(注)本題前可職女の下方法解之
作△ABC的外接园。作OD2BC.則:
BD=CD==BC (143474 X+9)
M 8C=ABTAC-2×ABI ACCK108*
5+4 + 2 x 5 x 4 cor: 72°
=25+16+60 x 0.3096
53.36
BC=153.36—7,305.•
/ BOC == 2<BAC = 216′′ <.800-366-216=144
因為等腰三角形底边上恶
OR = COI L BOD
3D
OD = BD CTL BOD:
- 1182
11. me adsorption of water in sp11 by the root
hairs is due to the (1) diffusion (2) osmosás (3) transpiration (4) root pressure 15) capillarity
11
4 substance evolved during photosynthesia ie called (1) starch (2) carbohydrate (3) water rapour (4) oxygen (5) carbon dioxide
12.
An instrument used for measuring the rate of transpiration in a plant is called (1) hydrometer (2) thermometer (3) potometer (4) clinostat 45) a levar auxanometer
13.
A process by which living organisms produce energy from the breaking down of food in the absence of free oxygen called the (1) oxidatim of carbohydrate (2) aerobic, respiration (3) essi letion (4) liberation of energy (5) anaerobicsiration
16.
15. Respiration in a plant involves an exchange of
gases. The gaseous exchange is invisible. because it is by diffusion through (1) stomata (2) epidermis (3) Mesophyll (4) vascular tissue
5) spongy tissue.
15.
=3.653 cot 72°
LOG
(as)
0.5627
(113) =) 12 α, p 3 ax
@AH=210D = 2,374 (uf)
cot 72°
75118
△心顶点之距離等 ・距離之橋)
0.0345
_ath+c= b+c) cos A+ (c+a) cos B + (a+d) cos C
fra
(AE) B
18.00
#$29
分解
Cos 8 + Cas A
(= basc + c cas B
b
acos C+C ces A-
bra={fta) cos C+ C(CH B +00! A)
(b+R)- (b+a) cos (= c(com 8+ cor A) (b+c) (1-cos c) = c(cosB+ cos A)
and B+C A
(4) ABCD 3 ZA‡Z ALF} %.
DA=6 ·求LADC的度數
(18) £1 bo to []; BAC ADC
AC == 6 + 7-28687 CASA
= 85-84 cot
因园之内楼四边形對角相
* <ABC-180°-0.
△ABC中,依飾法定律
AC=5+4-225 14 001 (186°– A)
cul+40 cat 8L
4/+400+0=85-34088
(£)
3=5 BC=4CD=
12.400$ A == 44
3548
TADC=69713:
✡★ 13 8-67° 13°
(5) 3®œ#{ ABCD AB=5 BC=3 CD=3, DA=4
AC=6 求BD之長至三位有效數字.
(解)在△ABC中传餘弦定律
cos a =
查表得
LA A ADC
--#3=0.8958
LEADH Quetta.
•$6.16
(4-) 32 2 8 B
(a) x2+ß3
( 18 ik –) 10 18 * 15 1 = M 1}$} &+8=-& BL
-zac
x+p} (a=d8+p3)=(a+f) [(a + B)2= 3xß}
[(~£)-3---
завство
b-zac
(解法=)因。,為原万程式一根以之代入
apr
a{d+p3)+b(a+ß)+ ZC=
4}=} @(+83) = - b (x+ß) — 2C
以栗 (1)弐
相加
4324
b(£) - zc
Rac
a(x2+8)+ b (x+8)+C(x+8)== al
〆為根て方程式 (解法)因十月:
方程式で両根和
ザー(西根和)y-(西根付)
-(1-aac) y +
两根精
值
fx_b_22c_c(£) [4x]
122
заве-
- braafe be sube-t'
az
Sabe-l
—
safe be
(解法二)因y=x. 代入焼方程式
自案
(1)謝の日
a(±17)+b(±/7)+
ay±by+c=0
=-(ag+c)
by
by=ay2+2acy+c
•b)y+c2
第四次預習題:
зас
-3=0的两根,求下式之值:
•*. (b) (a+b)(A+Z).
(2) 31 × 83 5x+2x-4=08** * 17 311 12 1
the (a) 2x+p ap+x, (b) 2+1 3+1
(3) 3) x-(1-2)x−(m+3)=0 ££££ = √ #| ‡ 14 3
25 求m之值,並解此才程式以核鲶三
(4) 31 m, n B †12 & Xie 3mx={2m+ 3n}x+2x=0
之两根亦為有理數
(5) 3| 2x-3x-3+R(X+X+2)=0 = A*B **
求值并核驗之 (6)
sam z 11
(7)設
b c 1 720
-C) + (x−c){ x-a)
#f #11 ab-c
两根 同值而異號