1969-04-04 — Page 27

REFERENCE

育教僑華

HALL

頁三第張七第二日八十月二年西巴夏

VAI KIU YAT PO

三九六九市中文学金試題預習

(C) sinx-cos x = 0.6

數學科(二十二 )

*喬仲强

Bosx

sta, zanz 化簡

第廿一次預習題解答

tan +5 tant

(a) 5c010+2 sinė = 2.

2x/

=0.7015 -5.7015.

(1)解下列三角方程式,答案由-180° 至+180

(M) A core 1-sino, uz 4 X 817 sinf z=XZ 程式

5()− sin^8)+2sin0—2

去括号移項化簡 分解

自

* A

查表得

5 sin 0-2 sino- 3:

(4) ₺ x−−143*8, Ax

-0.5(不满足)

(i) £ tan2 = 0.7015, ==72×180°

m_x=xx 360*+70°6′ (LÃ11)

巩!時則不合題意 2=0, 91 x=70°6

ten 2 =-5.7015, 3- nr 180 - 80° 3′

X=Nx 360° – 160°6′ (LAX). 2004, 7t, 2-1. §] x=179" 54"

(a) LB 11 X=nx 360°+70°6" } Nx360°

- 70°6′ 3X 199°54

度

(解法二)乘以吉

♬ sin 45° = cos 45° ——

sin x cri 45°– con X di sin (X-45°) = 3√2

X-450 — 7 x 180° +

=nx 180°+ 45° +C

10

x=706

X=199°54

-160°6

0.6 – 3√2

0.4243.

°6′ (111)

X=430°6" (A

(sin 0-1) (5 sin.8+3):

1=0, sin 8-), To=90

(4) 5sin8+3=0, 8] sin 8=-0.6

而

36° 52 -14308!

70-90° 36°52′ 34-143.8.

(b) tan x + sec X=2.

(解)移項

sec X=2— tanx

=4-4 tan x + tanx. 4tanx=3

tan x-3=0.75, 36.52 34-143°8.

驗筽:(优)若x=36°52′,代入原方程式

stan 36°52′+ see 36°52′

07500+ 1.2500=2(142)

#25 = tan (~143°8′)+ sec(~143°8'")

= 0.7500-1.2500—

☆ x=36:52

(謎)因解此方程式時経過自築手續故別入根

(31)94) 18 seex-tanx=1

$4(2),(3) T4 £34** 19 tan x= seex-5

故能適合原方程式的,祇有第一象限角

Tel + 2 uno cos 0 = 3 cos2 8.

(17) 44-4

An Oh and cose-300400.

(ain &-2018) ( sin 8+ 3 cor: 01-0.

@] sino = cos 0

(i) % sino-cos A=0,

4 tano=

kk tan

答:

8-45° 36-135.

Jain -3 cos 8.

0=-71°34 34 -251°34

0=45° -71°34", -135° 3 -251934.

(d) 2 cos20=1+sing

(46) us ( 1 − sin 0) *8 4 cos 0,

x=0.3403, §] x= 70°• 但289°54驗算不合

(u) = cos x=-0.94603, x= 180 19

相除

分解

分解

seex + tanX:

secx- tan:

2 sino-1= 0, 0} sint-

nx180°+46 +(-1)*x25°6′

赚

谷(通値

(b) X=70°6′ 3 199°54

==(1) [恒等式」

-(2) 【原方程式]

(3)

(解法⇒全式第以5

移項 自案

5 simx-5 cos x=3

5 sin2=500sx + 3.

25 mx (523)

25 (1− c1x)= 25c1x+30cmx+9 42 AL. 25 cos x + 15 cosx-8=

15 ± 152 4X25X(8)

-15±5√41

x=199:54

0.3403

$ = (a) i} } x=n

2x25

-0.9403

5*6(驗算不合)

(b) x=70°6′ = 199° 54′

15± 10.25

6:401

(3) 用图解法:解下列方程式「答案由0°至90) (a) tan x + 2 cot x = 34.

2(1–sin^0)—14 sind.

2 unė + vino

(2 un0−1)(sin8+1)=0

·0=303150

X

tanx

O

cot x

OF

sin &+ 1 = 0, 9] aing

答:

£ 30°

(e) 2 sec10=5tan 0

(04) 3] see 0=1+ tano

2(1+ tan:8)

Stan 0.

Ich stano-5tan 0+2=0

・分解

(tano-2) (stano -1)=

(4) to tano-2=0, §] tant=2

63°26 9-116°34′

(W) X 2 tano-1=0. Of tan o

9- 2634 3-153:26.

• 0=26°34 63.26-116.34"

・0=-90°

m153°26

(2)解下列三角方程式,先求其通值,再求0°至360°之急

(解)分解因式 (tan (tan 15) 1

(i) t tan x-1=0, § 1 tan x=1

x=nx180°+45° (1448)

雷 n=0,則xX=45

x=405° (7421).

(u) ‡"tanx¬√3=0, Q1 tan x = √3

x=nx 180°+ 60° (In {L)

n=0, | #4, #1 x=60°, 240*

{x: (a) il 11 X= Nx180°+45° = nx180°

(b)

(解)

() = 45°, 60°, 225° & 240′′

8

A cos x = (1- sin x).

44-445-

代入

去分母

分解.

x+1-24x+

2 sin x-2 cinx +

(4 sin2x-1) (4sin x −3)

-10 f sin

x180° ±30° (104ÍL)

x=±30° ($379202) X=210° 3 150° x=390°(不合题意)或33

3=0, §] sin x = ±±

x=2x180° ±60°·

當n=0時則

x=±600(夏号不合意)

X=240° 120°.

2 x=420° (74) & 300°

(b) x=30° 60° 120°, 150, 210, 240, 360°

330

(OR) y=tanx + 2 cotx° = [] # *****: 0° 70° 20° 30° 40° 50° 60° 70° 80° 90°

0.176 0364 0.577 0839 1192 1.732 2.748 567)

5.671 2.748 1732 1:172.0.839 0.377 0.364 0176

:00 11:52 5,86 404 3.22 2.87 2.89 3.48 602 00

Jou & $ (x-10, y=11.52134*4*). * ALLI

24 y = tanx + 2cctx 14 At A- B+ bess

直线,得交奌之横坐標即為巧求之根

答:

X=33

33 ·72.5.

=34

60

in (3x°+40°)

(b) sin(x2+10°) ==

(解)先製成下列两表

000&+wwitem

X-72-5

y === sin (x°+10°).

440° 0° 10° 20° 30° 50° 60° 70°. 80° 900.

20° THE 100

30° 40° 50° 60°

80° 90° 1000. 70° 40.174 0.342 0.500 0.643|0.766 0:866 0.940 0.985 1,000 0,985

=sin(3x2+ 40°)

0°

·10° 20° 30° 40° 50° 60° 70° | 80° 90°

3x2+40 40°

·70° 1000-130° | 160′′ | 190° 220° 260° 280′ 3/05 y 0.643 0.940 0.985 0.766 0:342–174 - 643-940-985-766 然後分别绘出两曲线y=sin(x+100)及

y==sin(3X°+40°) *‡ ‡ 3 § ≥ ƒ☀ ☀ ** By Z B *.

五期星日四月四年九六九一曆公年八十五國民華中

(接第七張第二頁)

生物科 (H)

reconverted into glucose and carried to the body by means of blood circulation.

(iv) The deamination of excess amino-acids to.

urea to be excreted - Host of the amino-acida are used for the synthesis of proteins of

the body. A small amount of amino-acids may be stored in the liver, they can be decomposed

which into harmful wastes such se aumonia,

is, therefore, combined with carbon dioxide. to form the less toxic urea and is excreted by the kidney, N

(v) For conservation of iron formed from the

breaking down of red blood corpuscles - The iron forming from the haemoglobin of the destroyed red blood corpuscles is extracted by the liver for use in building up the new blood cells.

(b) In man, there are three pairs of salivary glands They secrete a solution, the saliva, of mein and ptyalin. The water of saliva will dissolve any soluble food, helping it passes from the oesophagus to the stomach, while the ptyalin serves as an enzyme to digest the starch to maltose.

The chief functions of the stomach are the storage and mechanical churning of food, and the Initiation of the chemical breakdown of proteins. During digestion, the gastric juice is secreted by glands lying in the wall of the stomach. It is composed of water, mucin, salts, hydrochloric acid and enzymes, the pepsin and rennín,

The hydrochloric acid acidifies the contents an the stomach, stopping the action of the ptyalin, activating the pepsin and rennin to function, and killing part of bacteria present in the ingested matter.

The pepsin digests protein to peptones and proteoses, while the rennin is responsible to coagulate the specific protein, the milk. By this, the milk stays in the stomach longer and the action of pepsin upon the milk is ensured.

The pancreas is an important digestive gland, producing quantities of enzymes that act upon carbohydrates, proteins and fats.

The pancreatic juice contains amylase which further digests starch to maltose; trypsin which breaks down the proteins into amino-acids; steapsin (lipase) which converte fats to fatty acids and glycerol.

Questions for this week:

1. What do you understand by "Respiration"?.

What is the difference between external and internal

(tissue) respiration? Explain. the importance of respiration in the lives of living organisms.

Describe the structures of the lungs and trachsa of

Named mammal. Explain as far as you can how this mammal fills its lungs with air.

X=32.5

32.5

第廿二次預習題

(1)分解下列各式為因式一

(a) (a+b)(b+c) (c+a) +abc,

(b) (1a) (1+b2)~(i+f)2 (1+α).

(C) 45x-27xy−28x+6y+4.

(d) (x-2y)x2- (y-2x) y3

(2){a}有两但二次式,其H.C.Fx1,x.c.m.骂

X-7x+6. #wh that.

(b) A * * 100 = K #H.CF. B 48, L.C.M. B 1008,

求此二數

(3)甲校男女生之比马32:31,乙校男女王乙比

5:4.若将两校合併後,則男女生之比為89:8元. 求甲乙两校學生送數之比,

(4)某船航行相距18里之两鈕間,設順流速度 ̇較逆流速度每時多10里,又逆流航行較順流航 行多费54分鐘,求船在静水中航行速度及水流

·速度,又此船往返之平均速度為每時若干里?

(5)色知一级數之首儿項之和為52,試証此級

數蓴等差級數,並求此报数

16) sk a, b, c à 14 122 à

log (a+c)+ log (a+c−2b) — 2log (a–c)

(7)(a)$/= y=x+÷x<««' (x®116-22+2)

(8)利用此图线以解方程式

「答案須準確至小數一位)