1969-03-27 — Page 23

育教育三第張內第日十初月一年西己展會

WAH KILI YAT PO 郭日橋

一九六九本中文学会顯969 文學考題習

數學科 (廿一)

*喬冲强...

生物科

(f)

廖百琴

BIOLOGY (21)

第二十次預習題解答

1) A,B葛直线XY 同側两定奌,求在XY上取奌c

MAR LACX=1BCY

Eto $14 ko ki sha)

(1718) | 12 A 47 AD LXY*

31 & ± 2 A' AR DA'— AD

2 TË AB 2 XY ✯ CA.

#CAR

(証明)讀者可証明

LACD=ZACD= LBC

▲ ADC SADC 12 13

(註),此題為利用A算關於KY之对精

2 ACTBC $h

12) PA ILA = D={ ABCR ABLE I 2 * £

BC CAZZ & Q R

< ARP=2CRQ

Zła ★ 14 ka 2383)

(1412) | L3 PIE PDLBC#

延長之至K使DK=PD. H.

15 <P Q B =

Answer to the questions of last weekti

1. Make a drawing to show the structure of a Named tootl

of a Named mammal and describe the functions performed by the various types of teeth in that mamma 1,

ens.

This drawing below is a vertical section human molar tooth

1十十月三年九六九一番公年八十五福民華中

purposes:

To support the soft parts of the body, thus,

the definited shape of the body can be maintained.

(ii) To protest the underlying tissues and vitaz

organs of the body.

(111) To provide an anchorage for the muscles. (iv) The red marrow performs the productive function of red blood corpuscles.

(v) To provide a levar system on which the

mascles act, thua the movements of the bodr are carried out.

Coracoid

process

Crown

Enamel

R 2

··Denline

Hume

Gum

Neck

Glenoid

-pulp cavity

Head of

cavity

Humerus

Cement

Kadius

Root

-Alveola.

membrani

•Supra

scapula

Spine of

scapula

Olecranon

process "Sigmoid" "Notch

Trochlea

Ulna

ZHIRCH=1E

212 P17 PEL CA ISE

3 12 HKBC CAS

* to Q R BY 344

(3)拾定直线×Y 同侧有定园口及定奌P求於Y上 -奌C使直线CP及由C至0园之切貸典XY成等角

已知)直线 XY同側有0.

(44) ×Y + a CR 1£

CP及由C至0园之切线”

與XY成等角

(18) La P. 14 PALI

XY并延長之至Q奌

112 AQ=PA

2 18 Q 14 0 E 2 m 14 QB, a

QB2 = 2 x Y at C, B C2

3C及C均為好求.

(4)一定点 4介於两已知园 0.0之間求過A作.

线典两园台支拎 B C 而使 BA=AC

(已知)O·O两园

BAR

TEIDO AMFI

线 BC备两

1 * BBCM.

42 BA-AC.

(作法))联 OA并延長之

ZPIZ AP-0, A.

以P笃心, 0园之半径笃半径

14 0 2 0. 1 KC, & C.

3 TE CA (~1.2) † 1 § ÷ *o Ak6;

4 B, C, B B, C, DJ Žπ *

联 PC, 并通 0, 作英平行线交GB(或其延线)

於D則因40=<P(平行线之内錯角)

< PAC, = <0, AD (NTIA)

3 APÁC,200,AD ( A S A ♬ 0, A = AP171)

0,D=PC, = 0, @ #1 € ( H÷2)

D贞在因上(-吳奖团心距離等於半径)

DIAB§ (@ 6, 5 CA) 0, 8 z à „§ 钽因 DAAC(上證全等As之对石边)

A** ** 8.A=AC_

(註)以A吳尊対稱中心,則?因為0,园之对稍形 為如果將0國依A貞旋轉180°则0园重合於ㄗ团 此種对辑,在作图题时常利用之

5) 0・為直线xY同侧两走园,求作-直线界x丫平 行而各支两圆挂AB两便 AB等於定長 a

Blood vessel. and Nerve fibres

A Human Molar Tooth (L.S.).

A representative molar tooth consists of a brown projecting above the gum and two or more roots embedded in sockets in the jaw. The crown covered by a layer of enamel which is the hardest substance in the body and consists almost entirely of crystals of calcium salta. The rest of the tooth La composed of dentine, a substance very similar to the bone. In the centre of the tooth is a pulp cavity containing blood vessels and nerves. ▲ layer of cement covers mich of

tooth firmly in place in the root and helds the

the jaw

The dentition of mann le is adapted for the purpose of obtaining and handling a wide variety or foods. Their teeth are not all the same shape as is generally the case in man. The human teeth are differentiated into various types. Chisel-shapeq incisors are present at the front of each jaw and are used for nipping and cropping. Next to the incisor is a single canine tooth, which is primitively a long, sharp tooth, useful in attacking and stabbing the prey. A series of premolars and molars follow the cabine These teeth tear, crush and grind up the food. In addition, most mammal brezic food up mechanically with their teeth and mix it wit saliva, which usually contains an enzyme that begins the digestion of carbohydratan

2. (a) khat are the functions of the skeletons in &

AA1? Give a labelled drawing of the skeletons of a fore-limb in a rabbit, Describe carefully. how its movements are carried out. Describe briefly the following types of bone Joints: (1) a hinge joint, (ii) a ball and socket, and (111) a sliding joint, giving two examples of each, in the rabbit's skeleton.

a) The skeleton of manual serves the following min

(314) 318 <P,C,D, INTE, AP, P.D, Y TE at ▲ AP.C. Irt AD, P, C, AT TATL

(7)在已知角xDY 内有一定奌P 求遏P作=密直之直待

(Z fo) P G Z A LOY

(*#34 ) 12 P 44 = | # /* 5 *

OX, OY IS AB B KO 12 PAL PI

A PA=18.

phalanges

Metacarpals

Carpals

fore-Limb Of Rabbit

The movements of the fore-limb of a rabbit are carried out mainly by the alternatively contraction and extension of two different groups of muscles, the biceps muscle and tricepa muscle, When the biceps muscle contracts the forearm is drawn up, hence the lower end of muscle marks its insertion, whilst the end attached to the shoulder-blade, which does not move, in its origin. To restore the fore-arm to its original position another muscle is brought into play. This muscle is called triceps muscle, which is connected to the sans two bones as the biceps, but passes behind the elbow joint. When it contracts the arm is straightened out.

(1) A hinge 1oint: it allows movement in ons

plane only, e.g. knee and elbow joints, (11) A ball and socket joint ; it allëws movement

in any plane or direction. e.g. hip and shoulder joints.

(111) A sliding joint: it only permits two bons. surfaces to glide one over the other.ro.g. most of the joints in the wrist and the ankle.

Chiestions for this week z

1. (a) What is digestion and why is it necessary?

in a large scale make a fully labelled drawing to show the structure of the alimentary canal and the associated glands in a Named mazuren: 1. and explain, how the structure of the alimentary

of this illustrated mammal enables it to carry out its physical functions,

State as fully as you can, the functions of the liver of a mammal.

Explain briefly, how do the salivary glands, the

stomach and the pancreas help in the digestion of food in the body of a manual.

(A1) | 12 AB

Le 014 00 + AB # NE § 2 2 0 12 Co-oc

以加為直径作半园 只以口为心, Â為半径作0园:

為心,如為半程作弧半园拎D..

5联 OD, 並遇 A作英平行线支oo两园拎P及P

12 8 14.00 ÷ 9102 »B 10AQ

7 AP, 8Q $4 *

已知)O,O、為XY同側两定园.a笃定長

(求作) -直线典XY平行而交两因拎 A及BB令 AB=u

(14) 1.13 0, 170, PH1 XY, £ 4 0,P=a______

之以P為心,0园半径尊半经作园交Q园1B及B

3 A PB1 (i = 2) # 180, 14 $ #15 1 20,8€ A;

4K A.B. (<= 1,2) By By My Je

[6) 12 @ + - 6** §. 1244 – 3 12 $##<89=15. 已知)口為两同心园之园心,A笃

外因周上一定兵

本作)過 A作外园之 AB

使其笃内因而三等分

作法))以A笃心, 内因

ż直徑為半径作弧交

DETT, AP.

联PO并延長之要内园

P.

3 T AP, † KEZ < 24 @ 78,

1 13 2 2 3 3 12 14 AB,

5 AB, & AB, 4 *

(1/40) | 15 P47 PC LOX.

Z 15 P1] PC 2 & 13 D. D,

3 NXP BEC PC Ž ¥4443A 2. D. D. # D, & D, FAR

4 1§ D. (1−1, 2) 14 PD, 2*** Z OY * B.

5 74 BP #1574 £ £1 & 0* * A.

PA: 2 PB. Ž

8)遇口因上两定奥 AB 求作两年行弦使其和等於 定長。

(註)如以為园心如為半徑作之弧交半团

3.8备作0D×平行线亦為碚求 ☆ OD(ka)小於06 本题有两解(图中未给出)

第十一次預習題

解下列三角方程式 答案由-180°至+180°

(a) 5cm*8+2 an8=2

(f) tanx + MCX=2

(c) un0+ 2 amb utē = 30*0

(d) 2 cos

(e) 2 seco

tano

解下到三角方程式,先求其通值 再求: 0°至360元角

(a) tanx-(1+13) tan x + B

(C) smx Cod X min 0.6

3用图解法解下列方程式 答案由0°至900

(a) tanx't sutx=37

(f) sin (X2 +70′′) == sen