1969-03-13 — Page 26

頁二第張七第二日五廿月正年西己満夏 WAH KIU YAT PG

一九六九丰中文中学會考試題預習

(註)由正法定律

abc

又由餘弦定律,C8A=

數學科

(十九)

喬仲强®

1第十八次預習題解答

COSA: cas¦: co C—25:19 2.7.

(1) SABC+, sinĄ: sin B: sinc=5:6: 7. Ex £1

DinB

sin C

-5:6=7

alg=t=

-k. §. a=5k b=6k, c-7k

(68) + (78)--(SR)

報日橋

· ABD ==× 100×60 amdo = =2954 (34)

3000X 0.9848

ABCD = 2954 joy + 2488 sof

5442Ju

AD

A CBD=1×100X 50 un 84' 21'=2500 × 0.9952

=2488 (Fof).

(5)差 AD為AABC之分角线,試証

(証)設如右图:在AACD中,

1

四期星白三十月三年九六九一曆公年八十五國民華中育教儒

ま

VZÁS #2%$#$%I SIGFIEL

*#$%3 #3%$ #3%3 #$%$ #853 #90**#3%3A

依正弦定律

7 A ABC +

(((2) 74(1),

2 x 6 R XF R

sint

[coB= Rea

2 × 7 1⁄2 × 5 k

gif gif gbsin A

sin A

aa

(2) // seco=3

(A) @ tac 6 — sec 0,

(~ + =)2-1

(

(為求正

2m

•

[cos(= a+f=c2 __ (5R)+(6R)2= ( 7k)* ___

cos A+ cos B: cos Cam 235526:19 3.7

tmxsico – tano 2 11 že ho ?.

19

seco + tan 0 = m

besinA (p+g) = ta(b+c) sind.

besin A-

been A

W AD-

t(b+c) sing

bc sin A

(x+c) sin

Q.E.D.

seco - tano

(3) × b box (4)x C. 180 p+q=a=0,

(b+c) und

(6)一正四角錐体中底每边5吋正方形,高6吋試

求其(2)表面積 (6)一楼典底之傾角

【解) 因V-ABCD 為正凹角錐,故高线必

14 ABCD = Pic Of

OE=2x512 = སྒྱུ་

VE−√62 + ( =) — —

AVAB

* ABXVG

=女×5×号一號(吋)

Z ABCD

= 25 (2)

1969 PD13

生物科

十九)

慶台灣

BIOLOGY (19)

Answer to the questions of last week:

1. Make a fully labelled drawing of the skin of

mammal as seen in vertical section and give an account of the function of its various components.

Ans.

The diagram below is a vertical section of the human skins

Malpighian layer

Sweat / porem

skin recep

for

Epidermist

Homy..

layer

Granular

layer

Dermis

Blood

Vessels

Sebaceous

赤或m (答)

(3) △ABC中,若a, &及A為已知,問在何種情況 下,此三角形爲两解题?又器C 及C,為此两倡三 角形第三边,試証c.

(答)△ABC為两解题時,其必要條件為2A<90°及

bain A.

(31) Elken to 13: AABC / AAB.C

$5 * = 1/2 = 3*2 QJB BC=BC=Q.

Leif CDLAB,, §l co=bant,

£ B, D = DB, AD=FEMA.

AB,+AB,

=(AD+ DB, ) + (AD-BD)

RAD [§ 8,D=28,]

abcos A.

(BHB) / A AB,C 4

AAB,C

a2=b+c2-2bc, cos A.

=f2+cz-abc2 COSA.

比较上列两式得

a b c

cos A-2bc, cos A

表面積4x

tan

-1.6970

∙AO.

★* 18 0-59° 29′

答:(Q)表面積90方吋(8)積奖底成59°29′之魚 (7) 1⁄2 tano + tand = p.

tan ord

x

cos O

以之除上式得

54F, (c+c2)(c-c1) = 2b (c,− c ) cos A.

JGCOSA

(4)(a)ABCD為任意四边形,对角线之交角茑日,

四四边形時,其结果則差何?

BASE 2013 LACKED sino.

(31) A ABE=1 AE » BE sin (1-0)

天器 ABCD 馬

== AE BE sin O

ABCE= BE CE anB

^ ABC= BE sing (AE+CE) = { AC+ BE sin 0.

MEZ AADC== ACEDE sing

ABCD == AC sino ( BE+DE)

ACY BD sin ✪

如ABCD為凹四边形時,則

AABC-AABE - ABCE

= AEX BE sin (7-8).

- 1⁄2 CE & BE ain (1-8)

== BEsino (AE÷CE)

£ AC× BE sin f

A ADC=& ACX DE sin 8.

4. ABCD == ACX BD sin t

Q.E.D.

(f) 四边形之对角线 尊7吋6吋,面積10方吋,問其过

角线之交角為若干度?

(解)由上面証得结果,

sin 939

查表得 6=28° 26

4x7x6ring.

0.4762

答:交角萬28°26'

(c) & OL#} ABCD+. % AB=60", BC=50 CD=

DA=100", 2A=80° A&tart

(解)在△ABD中,依餘弦定律

BD 60+ 100 - 2×60×100 cos 80°

ACBD中

|BD=50+100-2X 50 X 100 00 C

50+100-2×50X 100 COS!

60+100- 2× 60 × 100

2500-10000 cos C

3600-12000X0-1736

CH C0.0983

查表得 G=84° 21

(L) p2= tan© +tan 4 + 2 Iwn0 tand

> + sec2d + 2 pec 8 sec 6. (sec^0-tan ()+(sec24 – tan 4)

+2 sece seed –-2 tano tand

2+2× 30* CH4.

ඒ

2. Cos O cos & + 2 – 2 sing sind.

cos O cod &

2+2 (Cot§ cos ó-sino,

cos o cos

2[[+86(9+0)}·

asinler

9 cod♣ + cos O sing

cos & cos

sin(0+4)

1+603(0+4)

Cat A Los

[1+ cos(0++)]

sin a

將

第十九次預習題

D

(1)(a)何謂等差级数?又活於等差級數各项中乘以 (或加)同敷英结果仍成等差級數否?試証明之

(f) El a b c m 4 P. & a2(b+c), f(c+a) c2(a+b)

亦成 A-P.

(2)(a) T = {4 * A P. £40 3 18 £7343 126.44=0.

(8)有四数成A-P.其和為20,其平方和為120.求此四數

(3) 1⁄2 a, x, y, bmx AP, a, u. v. f / HP. (goda €). zxu af & x, y

yu al

XV

(4) bf a, b c AE

G.P.試話:

r & HP; ap bq cra

+-

(5)設

d musst az a+b b+c, 亦成等比级数

(6)在2奖9之間插入=數,使前三數成A.P.,後三數

成GP 試求此二數

(7)有一等比級數,其無窮項之和骂4.此级数各項 之立方的無窮項之和為192.求此级數

(8)某人欠債10000元訂明分十年攤還,年利率6%以 每年一期的複利計算,問每年年终应逞若干元? (9)某人購屋一間,即付20,000元,以後每年末付款

3,000元,十年付清:若照年利率8%複利計算(每年一期) 問此屋現價笃若干元?

Fat

I sweat gland

Nerve fibres

gland

Erector muscle Hair follicle

Hair papilla

The human body is covered completely by protective layer called the skin. The outer layer of the skin is thin and contains no blood vessels. This layer is known as the epidermis. Below the epidermis is a thick layer, the dermis. (a) The epidermis:

It is a thin outer layer composed of stratified epithelium. The cells in the deepest layer, the Malpighian layer, multiply and, in growing, push the older cells upward toward to the surface. As these move upward, they become flattened, squamous in shape.

The epidermis is again subdivided into two layars:

The horny layer

The more superficial cells of the epidermis, being far removed from the nutriment supply, gradually degenerate; as a result their proteins are transformed into a special type, known as keratin. It serves as a protective layer for the delicate underlying tissues from drying, from injury, and from the invasion of bacteria and small insects. (ii) alpinian laver

The deepest layer of the epidermis. The cells in this layer undergoes active division and so produce new living cells constituting a granular layer, as the formed cells move upward toward to the surface of the skin, they become flattened and die off, forming the horny layer.

b) The dermis

newly

The true skin, lies below the Malpighian Layer. It is formed of loose connective tissue,

containing fatty tissue, sweat glands, sebaceous glands, hair papillae, erector muscles, hair follicles, blood vessels merve fibres and sensory organa. (1) The hair

It is a hollow rod of dead, fibrous, epidermal cell surrounded by thin hard cuticle. The hair develops inside the he ir follicle, sach mir follicle is a hollow tube having a papilla-like structure at its bottom, from which the hair is supplied with nutriments.

(11) The sweat gland

It is distributed in the skin all over, the body. Each gland is a duct coiled up into a ball at its base, lying deeply in the dermis, and is supplied with a net-work of capillaries. The secretory cells of the giand absorb water with some salts, urea, and other waste products from the capillaries and pass them to the surface of the skin as sweat.-

(未完轉入第七張第三翼)