1969-12-19 — Page 26

=**+* -+-+RE WAH KIU YAT PO

報日僑華

五期:日九十月二十年九六九一寤公年八十五國民華中 育僑華

K% # %# $ % $ # $ % & # # %%# $ # 8 ;

#$%$ #1%$ #8%$#873;

$#$%$#$%$#$%8#$%$#$%$#$%58

#$%$#$%$#$%$#:

6=24°41′ 3 155°19′

•NO.

LOG.

#$%

Latπ, AC=51R, AB=AB=927

51

17076

46

1766286

¿CAB=24°41', <CAB=155°)¢

3.3704

AAB,C=AAB2C 9800K

980 2.9912

B#

3.37046

X0=2441 155°19′.

(6) ABCD 7***}, AB=10],

BC=bt, CD=6+ DA=8+, AC-125

求BD之長及此四边形之面積。

(解)在△ABC中,依餘弦炭律

2x12x6

罗僑

#$%#%$#

%$#$%$#

文中學會考試題復

考試題預習專欄

查得

數學科 (七)

喬仲强®

又在△ACD中,

12+6=8′′

2X12X6

第六次預習題解答

(1)(a) SABCY LABIŁA, TATE

a2b+c2-abc cosA.

(証設如右图,绳C作CDLAB;

並設 AD=P則

"God (180°~A)=- COSA.

p-bcos A

在TAACD中,依畢氏定理

#rt. ABCD a2=cD2+(c+p)

=b+c+2c (beosA)

v=b+c=2bccos A.

Q.ED.

(6)利用上题结果,或用其他方法証明 c=acmB+bes A..

()由上証 a=b+c2-2bc cos A.

同理可證

f=c+a-2cacos 8.

tabe, atba2+ b2+2c2 (2&c cos A+ 2 ca cosB)

#$1,120 = 2c (acos B+ &eosA)

(3 ) 3 1) š€ + + [ += bcos (180- A) ==LesA——

C=goo1B+cos A.

G.E.D

(2)

因2c=0,以之徐上域,

X # st ABCD + ++c=acos B.

(2)–(1), c=a cos B-(-bcos A)

若A,B均為銳角,(如右图示)

• cmacos B+ best A.

5] AD=&cos A, BD=a cos B,

(2) ABCD #wleπ * AB=4, BC=5; CD=6,

C=AD+BD=acos B+ bess A.

DA−7. * LABC 1973 32.

(解)联AC則在△ABC中,依餘弦样

AC=AB+BC-ZAB×8CCMB

-4+5-2×4×5000

= 41-40 cos 0

因ABCD 為园之內接四边形

+

LD=180° – 0.

在AADC中,依餘弦定律,

AC=6+72-2×6×7 cos (180°-0)

·85 +84 cos. B

85+84 cos 0=41-40cm 8.

移項化館400日

124

-31=-0.3548

← cos 69°13' = 0.3548 (E★#14)

*0=18069°13=110°47′

= ZABC=110 47.

(3)一塊边長為5吋,6吋,7吋的三角形木板,平放在桌

上,沿最長边旋轉300後,問其巧对奌,桌面距

離着干!!

(解)此三角板旋轉形成之

(dihedral angle)

E-BC-A (face angle)

# 30° # % ADIBC, EDLBC,

N

<EDA=30°,現在先求AD高线之長。

#QABC4, 5==(5+6+7)=9

AABC=19(9-5)(9-6)(9-7)= 4√6 Juf.

XE LABCO LADABC

ZADX)=656,

12/5

MAADE 4, - AE= ADain 30° — 12/6, ] -

答:距離為8吋 2.10 (f)

(註)因答三位有效数字取近似值時,不能21吋

(4) ABCD -7, ABIIDC, % AB=3,B C−4+

CD=8+, ZBCD=123°,✯ ADZE.

(解)過A作AEII8C 則 ABCE為平行

四边形, AE=4†, CE=31

70 ED=8-3=54.

X LAED=LC=123°

在AAED中依餘弦律-

·AETED ZAEX EDCH ZAED

4+5-284x5 cm 123°

16+25+40ces 57°

-41+40×0.5446

62.78

- AD=√62.78—7.923,

*Anv=7.924

(5)三角形的两边為51尺及92尺,面積為980方尺,求其夹角

(M) aka 0,41

A==×51×92 sim 0=980

980 51X46

92

查表得$=36° 20%

*

0.8056

ABCD * Bolt #{ art, Aj <BCD=0+0=92°35"

=<BCD-

ABCD =π

BD

=ain 28CD,

BC

2.50—246 sin 45°18′ = 2x6x 0.7230—868,

4 ABCD & M mik #}, 2] <BCD=8-4=19°55

BD=2X 6 sin(2x19°55')=12 sin 9°58′

-=12×0.1730=2.08, (2.078). ABC, S=2(10+12+6) — 14

X#AACD

s==(8+12+6)=13.

A ACD =√13 (13~8) (13-12) (13-6)—3455—21.33.

ABCD 面積: =29.93±21.33—511263 8.60 X: (a) 5<«l2*} a$, BD=8.68+, 70 $151.357;

(b) (y wiź#} G# BD=2.08+ 111 8.6071

7) A ABC 4, € sin Ą: sin B ; sin C = 2:3:4, cos (B+C) = cos(C+A) : cat (A+B) ZŁĄÍ.

• a:b:c-2:3:4,

設:===t,

을===t

9] a=2t, b=st, c=4t stark ***.

(3t)‍+(4t)-(2t)

cos A

cos B =

2a0

cat &==

a+b=c2 Rat

2x32x42

(2t)2+(41)-(3t)

2 x z tx 4t (2t)+(25)2-(4t)z

2x2tx32

21t2

ZA Cos(B+C)=cos ( 180°-A) === cos A==},

·cos (CTA) ==cos B=~//, con(A+B)—-

cos(B+C): cos(C+A) : cos(A+B)=(-) (-1) +

- 14:11:(-4). [#]

(8) (a)+ y = sinx + sin2x° 11 (XT 100445). (f)利用此图像求 sinx+ ninex=1之根(準確至古度) (M)***TE: 4= sinx + sin 22.

45*

59 -20° 15° £20°. 251-30° 35 40° nimx 0 2087 0.174 0.259 0.342 0,423 0.500 0,574 0.643|0.707 imax 0 0.174 0.342 0.500 0,643 0766 0.866 0.9.40 0.985 1.000. 0|0,36] 0.516 0.959 0.985|1.189|1366|1514 1628|1,707 描出各桌後,依次连结之,得图像(graph)如下巧示:

= sinx + sina

LO英文中學會考試題預習專欄

文書院主編

生物科

· BIOLOGY (7)

Answer to the questions of last week

(L)

1. What is meant by the term "photosynthesis'

Ano

於同一图中, 方程式之根

4=1直线,其交奌横坐標,即端开求

*; X=20.5.

Ana.

第七次預習題

(1) % 2x-3x-3+k(x2+x+2)=0 2# ### #kz 值,又其根為何?

(2)1⁄2 ax+abx+c=0&#*31* *

ax+z{e+b)x+a+ab+c=0 =M*****

(3)設x*+(3k+2)x+2k-5=0方程式中,若一根.

他根之三倍,試求龙之值,並核验之,

(4) £ ax+8x+c=04 px2+qx+1=0@1++, A−‡‡8A, SE (cp-ar)2=(br~cq) (ag-bp), (5)若x,p為ax+bx+c=0之两根,求下式之值:- (a) (ax+b) (as+f), (f) (bu+c)(bp+c). (c) (ax+b)2+ (as+f)-*`

(6)設xp為a8+28x+C=0之两根,求作以及

132 為根之方程式。

(7) = a+b+c=0 at E

a2+ + e + + (1 + + + 1) =0

Describe how you would demonstrate that (a) carbon dioxide and (b) light are necessary for photosynthesis?

Photosynthesis is a process in which certain carbohydrates are synthesized from carbon dioxide and water by chlorophyllous cells in the presence, of light. A summary equation representing the procean of photosynthesia is conventionally written as the following:

Radiant

600 + 60 Energy

Chlorophyll

(a) Experiment to show that carbon dioxide is

essential for photosynthesist

Caustic polash

Solution

starch-free plant has one of its. leaves Insepted into a bottle, in the botte of which there is a small of caustic potash The oustic potash will absorb the carbon dioxide from the air within the bottle. :

After exposure to the light for sone hours, the leaf will be found to have no starch in the part which, being inside the bottle, was deprived of carbon dioxide whii the exposed leaves contain, abundant starch, (b) Experiment to demonstrate that light is

necessary for photosynthesis?

Cotton-thread

Black Cardboard

paper

The apparatus is set up as shown in the abov diagram. Two strips of black cardboard paper are placed to coincide above and below of a leaf of the destarched plant.

The whole apparatus is then placed in light for a few hours. After which, the presence of starch can be shown in all the exposed parts of the leaf, but the part covered by the paper strips is still staron free. Since the presence of starch in

is an evidence of the occurrence of leat

photosynthesis, the experiment thus proves: that light is necessary for photosynthesis.

What do you understand by serobio and anaerobic respiration? Compare as far as you cam, the differences between respiration and photosynthesis.

(b) By what experiment would you show that heat

is given off during the process of respiration.

Aerobic respiration is an oxidation process which occurs in the presence of free oxygen. During this process, the intake oxygen combines with the carbon of the carbohydrate, so that the products of the process are carbon dioxide and water. In other words, there 28 an entire braakdown of the carbohydrate and complete liberation of energy. The process can be summarized as the

In anaerobic respiruation

602

Enzyme

+ Energy

occurs only

in the absence of a free oxygen supply, so the

the carbohydrate is broken down incompletely with only partial liberation of energy. The by-products of the process are carbon dioxide and ethyl alcohol. The process. of anaerobic respiration is represented he

an equation ast

12°6

Enzyme

200+ Energy

The following table is a comparison OL the differences between respiration and photosynthesist

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