第二第張四第
日一廿月三閏年午丙麿
「
肚香港
【會過有關問題,詩
貼上大會餘及上大
日常內平行,感大座 會,主要係將該會
裔日星期六下午 時军五時,在會
度一座談會,將於本月开
殷代表(聯絡
我師會本年度第二次
日只代表聯絡
大胎璧举作,演出者
【音壁晚會,分下午月
第學總會學組主辦之
「今音樂晚會
培正同學會
WAH, KIU YAT PO
談座員絡聯會師
覆囘早請表代加參
中中考試題預習專欄
c=
喬仲强 •
數學考
(十九)
第十八次預習試題解答
9.(a)已知直角三角形斜边與一腰之落及他腰之長,求作此 三角形
(巳知)t△ABC中,斜边典一腰之差,d(=c-a),他腰长
(求作)△ABC.
d.
(作法)人作t△ACD使ZC=te
AC=b CD=d
2.作 AD之中我待典DC延
3.联 AB.
4A ABC下求
QE
(别遇四边一边上一定点,引一直线分此形為二等積部分
(已知) P篤四边形ABCD 边AB上定奌:
(求作)過P引一直线分ABCD為二等積部分
(作法)1.联 PC PD
2.邊 A作 AF II PDECD 延线于F
3.遇8作 BE || PCDC延线於E:
4平分EF 柊Q.
5. 與PQ即為好求
Q. E. F
(註));本題之作法,是先变ABCD.
等積 ᅀPEF.
2. 若Q真不落在DC上時,
(如右图暗示)其作法应如何?
2.(2)設AB為B和园之直
C為回上任意奌遇C作切
线典 AB两桌之切线相交於E及F瞑BER.
BJ MCHAE
(設如題)
(証明)). AE1BF(同者直於AB)
41=22 23=44 (#15 h6Y3)
AAEN AFBM(两角等)。
AE — AM= ($818BAŽE)
FB
FM Z
{AE=cE__FB= FC (切律 等镸)
CE
(代换)
EM
7 MCIIAE (分三角形两边成比例之线典第三边飞行)
QE
(B)样刊的两底分别 8吋及15吋, -直线平行於其 且将样刊ś為西等積部分,試求此直线之長
(解)謢如右图,MN 笃 等分样广
形AB♡之左边平行线
并設 MN=分时
過DM分别作CB 之
A
15"
儒·水阿公立建湾明校
金就東主余之好補
行赴英,試約數月
韩及與冰長等職,渠
減習事業共人,任 二深造,張氏致力保
十日乖英國海外航
校長張斂財,於
• 在英國深造
【特〕界上
張毓昆校長
: bain C 4ain. 81°16"
sim B 5.910
[A123" 14則C=14°46
Isia G
Ain B
=15.24
= $sem. 14° 4£ ' Ain 4207
答 C= 5.910 15.244.
NO
三期星
·這項獎金,在
星 得無上的榮譽
·校凡有二個學生
家
期物化學學生,已
得理士學位,現
· 佘扰媒露該校:
LOG
aim 81°16" 7: 9950 e
0.5971
7 8255
0.7716
0.6021
pimit ub 7 4063 0.0084 7. 8955
(b)È4§¥ sin2 & tame + cos20 cot 0+2aino cos 0
X Fan A+ cots.
1.18*9/
(x) / 3 = sin's sing + corto. Caso + 2 sin BBOLD;
Coin*0+ cre*日*
Det  sín Ô
into
+
Cord☺ min
5.(a) △ABC中,証明
(正)由正弦定律
油餘弦定律 C A =
cos &
A
左方
...
tano + cot = to
tam A
A.E.D.
(b) & ABC+, SE4A (a+b) sin2 = + (a-b)on" ==c2
(BE) Aj = (x2+2ab+ b) sin £ + (Amzab+b2) cos" £ (譯)左方一
zab (co-)
+b2-zabcos C [
c2 (+3)
如讀者未習半角函數者,可用
*C - S(s-c)
Q.E.D
6.(a)不許用表,求or+on+an值
(AB) @ cos © sun & = = [am(0++) -
方式:
(+cm+cm)
== [(ain I sin =) + (vein 7 – am 27) + (sin 77
号]
]。
(
- sin 3) == 12 (-six)
答:英值第一章
(B)解方程式
sin x sin 3x=-
(38) ♬ sino sind
打线 DH及 MK. 則 △ANK OAMDH.
MA =*星?(因DHNC, MKBN 均為平行四边形
>*}#{ DMNC = £h, (x+8), *}#} MABN== h2(x+15)
故
=h, (x+8)= = h2(x+15)
12x=287
+ [one (0-4)=cœ (6+4)] 代入原方程式得 [2013-3-Cox(3x+2)]=
答:x=可
第十九次預習試題
Goth
化麗
答:此直线長 子时间
闪器
(註)若樣形两本為a.半時,則x
=0,則 cn2x=
3. ABCD 為团之内接四边形,DA,CB之延线交于E; ABDC
<* # * F (a) ABE BCFBAZ
G.
E, G, F
三奌共线()張EP. FQ為 ABC园卡切线則EP+FQC
或
設証如題
若证明(a)
J. 联BG, EG, FG
(两桌走一直线)
261-22, 23-24
《国内接四边形外角》
(国内接四边形对角互補)
21427=180*
(代换)
5, G, F 共线(西鄰角互補,外边成一直线) QED;
1.(a)每磅2先令6便士之甲種茶典每磅1先令4便土之乙種 茶混合後,以至磅之先令6便士之價售出,獲利252 周其滋、 合量之比差何?
(8)求下列各數之平方根
(4) Ibbul
(4) 4243600
艾一空心球半徑3吋,重28磅,此球質量出鲂呎 448磅、求其厚度
1)以1000元分奖甲乙两三人,其中乙甲多50%,两鞍乙 少25%,問各得若干?...
a,
led成连比例,求證 a+b+c+d
3.
証明
65EC (切线為制线及其國外线分之比 1. (EP2⇒ EB EC 1FQ*= FB FA
的設
c8.5cm 6Q-EF(国外一奖1割线,每一制线典
FB FA=FQFE 其园外线分之積相等)
P=EQ:EF(代换),
·EP+FQ=EG:EF+FGIF E=EF (EG+FG)=EF,
·FQ*=FG·FE.
(41200)
(解)由正弦定律
Q.E.D
4(a) $$ atf+ amo
證
(6)$ 3x+1+2 [2x2-x-6 之平方根
5. (2) a, lx2-10°C+17=0之根,求下式之值
(a) (Ita-a2)(14b-b)
4 (R) a ABC4, are 5 of, buget, abada' i cibża (b) xx xyt =Q, xy??b, xyz C, TEE
EXO:6691 -- 0.
A 56 46 或A125*14 (此值对拎△内角和
定理適合 11) * A=56°46' By C=81°16', beski
(g) log a+ (2-b)logo+(p-g) lug C
64 you 469- (x取值由-3至+3),並利用之 以解方程式 400-3x-9=0(答案須由图中看得準
日一十月五年六六九一圈公年五十五國民军中
K
中所之风索污漫中心與新聞學系臨合主辦之 ・〔特訊 ) 夯粥中文大衆人文學科研究
明日由吳灞陵主
報紙與誹謗
中文大學學術講座
華僑教育
余樹得爲港生季光
獲蒙大獎學金
在澳洲大學求學
九零四學術作有「選案之新閔第」 東少除及中醫院,主離新明 。凤氏之 日報。一九六至六三年間,先後任激裟 公報、大光鞭、中選民報、中報丶 年登富。曾任以下各怨之藝緝及主楣;唯江 : 吳氏從事新聞事幹, 有年所,經
,抵迎各界人士到校際游 北,用旁整講述,他爲下午王時至四妆, 華僑日報料幃見灝陸主,「部烁區誹謗」
日一生余橫得,以品舉發 涉,配各會盘类
其中会考試題預習專欄
化墨料
CHEMISTRY (15
(十九)
王錦釗
Q.1 (a) When zinc nitrate crystale are heated in a hard glase'
test-tube, they are converted into a mobile, colourless liquid as they dissolve in their own water of crystall-. isation. 1
Zn (NO3)2 + H2O
Zn (NO3)2+ 6H2O
The solution boils and clouds of steam are given off. On further heating, the zinc nitrate decomposes, brown, fumes of nitrogen dioxide (NOy) are given off, and when a glowing splint is put at the mouth of the test-1 tube, it is rekindled, showing that oxygen (0,) is also evolved. The residue, ginc oxide (ZnO), is a solid which is yellow when it is hot, but which becomes white when it is cold.
2 Zn (NO3)2
2 ZnO + ↳ NO2 + 02
A
(b) when a stream of dry chlorine is passed ovér a heated spiral of iron wire in a combustion tube, the wire glows and black crystals of ferric chloride (FeCl2) are collected in the cooler parts of the tube.
The green colour of the gas disappears, as the iron and chlorine, combine to form the chloride. 2 Fe + 3 C12
2 Fe Cl3
(c) When excess of carbon dioxide is passed into a strong
solution of sodium chloride saturated with ammonia gas, sodium bicarbonate (NaHCOz) will be precipitated as
white sludge. The carbon dioxide first reacts with the ammonia to form a colourless solution of ammoniшr bicarbonate e (NHHCO3)
募
+ H2O + CO2 NHL HCO 3
The ammonium bicarbonate then interacts with the sodium) chloride to form sodium bicarbonate, which is only sparingly soluble in brine.
NH, HCO3 * NaCl *
NH, C1+
NaHCO3 •
The
(d) When oxygen is passed slowly over heated charcoal in a
combustfon tube, the charcoal will burn brightly.. gas carbon monoxide (CO) which is formed will burn with a pale blue flame. The charcoal first combines with. the oxygen to form carbon dioxide.
C + 02. CD2
As this carbon dioxide passas over more heated charcoal, It is reduced to carbon monoxide
2 00
When white sugar crystals are warmed with concentrated sulphuric acid, the sugar turns yellow, then brown until finally a black mass carbon, is obtained. The concentrated. sulphuric acid, which is a strong denyarating: agent, removes the elements of water from the sugar, leaving a swollen mass of carbon.f
C12 H22 011 - 12 C + 11 H2 0
T: (removed by concentrated
sulphuric acid) 3.2'(a) Hydrogen sulphids is a weak dibasic acid, It will react:
with caustic soda solution to form two salts, (1) the normal salt, sodium sulphide (Na2S) and (1) the acid salt, sodium bisulphide (NaHS)
Take 400 c.c. of dilute caustic soda Bolution and divide it into two equal parts. Hydrogen sulphide is bubbled" Into one, thalf until it is saturated, if thia solution La crystallised out, white crystals of sodium bisulphide are obtained..
*NaOH + H2 S (volume)
Na HS colourless, solution of acid salt.
If the normal salt is required, the other 200 c.c. of the caustic soda golution is added to the colourless solution of the acid salt. When this solution is crystallised out. white crystals of sodium sulphide are obtained.
2 NaOH +Hg s 【2 volumes)
Naz S' + 2 H2 colourless. solution of normal salt
(b) Hydrogen sulphide wiil requce nitric acid to nitrogen
dioxide When hydrogen sulphide: is bubbled through concentrated nitric acid in a test-tube, the pale yellow solutdon becomes turbia due to the suspension of sulphur which is formed as a result of the oxidation of hydrogen sulphide, and a large volume of brown funes, nitrogen dioxide, are given off
5+
2 HNU, H2 S 2. Not
brown
Tumes
forms, a
in the solution (c) Hydrogen sulphide will precipitate many metal sulphides)
from solutions of their salta. For example, when hydrogen sulphide is bubbled into a blue solution of copper sulphate, a black precipitate of copper sulphide
will be formed, and the resulting solution is colourless
Cu SOL
H2 S
blue solution
.NH, NÓI NaOHC
L4+k+h+48 ́→
-70 gms.
Cu St+H230元
black Dpt.
dalourless) solution
HaNO, LẠNH
22+4 litres at N.T:P.
It. of dry NH, NO reeded to produce 22-4 litres of NHz at"
ANS
170 gms.
N2ot + 2 H2O
22.4 litros at N.T.P.
N.T.P.
忘 10 LiteS
70 gms.
10 x 70)gns,
22.4
31.25 8.
olume of N20 at N.T.P. that would be obtained from 70 gas NH4NO3.
22.4 Litres
31.25 gus
x 22.4)11tr
10 litres