育教僑 頁一第張六第
1972罡文中學會考試題預習專澜∗文♂은 처형을
數學科
BATANEIL WAH KIU YAT PO
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二期星日二月一十年一七九一曆公年十六國民華中
1972
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中文中學會考試題預習專利
世界歷史科
·歐陽鋊豪:
世界不史科預習录及答匀,首术(十一)第二日,明(1972)年至北润,仍些展第二道 沒判出。裝1972年香港中文中學會考騙啡,本科會考範细说,对生,五金
denominator by cose
1994.
MATHEMATICS (1)】
This course provides the candidates of
H.K.C.E.E. 1972 a general revision on mathematics (Syllabus A) and it begins with trigonometry.
Consider a
where angled
triangle ABC.
90 and let / ABC = 6. the following ratios:
AC
Then we
side
sin e
hypotenuse
cos e
adjacent side
BC
hypotenuse
AB
tán G
Opposite side
adjacent side
adjacent side
opposite side
sec =
hypotenuse adjacent side.
AB
BC
hypotenuse
Cosec = Opposite side
It should be noted that: I
(1) The abové trigonometrical ratios of an
angle are numerical quantities.
(11) The sine and cosine of an angle can
never be greater than unity; the secant and cosecant of an angle can never be less than unity, while the tangent and cotangent are unlimited between any values.
cos e
cose
Multiplying both the numerator and
cose (1 + sine) = (1 - sine)
cose (sine + cose - 1·
cose (1 + sine) (1+sine)(1-sine)
cosỹ (sinë + cose –
(1+sine) (cose+s
cosə ('sine + cas
1 + sine
Co se
R.H.S.I
Trigonomettic ratios of the angles
60°
90
30
The most importa nt trigonometric ratios are sine cosine and tangents since the remaining three can be obtained from them taking the correspording reciprocals. (1) The angle o
Consider AFOR right angled at Rand / PQR
ewhere e is small.
As 9 tends to 0, Pand' R tend to coincide i.e. PR tends to zero,hand length of FO approaches. Than of OR.
FR
tane
sino
tano
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ATED: 1917-191524 NOT Year
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(3) $1504)
乙:
TRANZEICHEN
·共佔40分,考生必須心。
詳細問15屆(甲、乙、丙护然签者方裂,比間佔60分。
世界歷史科預習題目:
503641783-1911).
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3. HARD, UTAE SETS (7)
A
LORD NAPIER (G) LEE, UCUZAB,
CBX
CF)
002 GO
sin o
cosec
tan e
cos e =
sin 9. cos e
sea
Again, let us refer to the above Fythagoras' theorem, we have
CB2
AC
Dividing throughout by AB
Langle,
= 1 1.6. (B) + (AC)2 =
(2) The angles 30° and 60°
Consider the equilaterál triangle POR of side "a' and draw PM perpendicular to QR.
From geometry we claim that M-is the mid-point of
QR and PM bisects OPR
/ QPM = / RPM = 30°
By Pythagoras' theor
=PM OM?
PM
(a)?
Co
Dividing (1) throughout by
tan 0+1=sec
cos e
COS.
Dividing (1) throughout by sin
sine
COS
sine
Now consider
sin6o
POM
co 360°
Jie 1st cote = cosec
Example (1)
Given see 0 =
tan6o
Find the remaining
sin30°
COM
trigonometrical ratios in terms
Solution: By Pythagorus theorem,
cos30°
(m2
(2m)
we have
外事
WALEEYYA) ROBERT MORRISON (B)H.
IS MATTEO RICCI (C) TILINON (DY NICOLAUS LONGOBAR
DI
D.
3. -
E LORD MACARTNEY (D)
MALORD AMHERST (B) BUE GEORG CHARLES ELLIOT (D) LORD NA
JAMES LEGGE
JUCHITARR : (A) WASIO.B. ROHINSON CRY
C JON POVRING (D) HENRY FOTTINGER
STRANXARXARO TALES MANIA -
All the above results are summarized by: the following table.
Anyle
Sine
Cosine.
tangent
Example (3).
Find the value
olution.
sec45
30
459
GO
LIN HIG
નાના ન
احب ايه
√3
cos 309
sin 600
cos30 = √3
sin60o
=
P:
ö
the above expressio
4 (4) a - 4 (√2)2 - ¦ (£) 2
adjacent:
cose
hypot
sine
tenuse
opposite side hypotenuse.
[cosece
tane
sin e
Opposite side adjacent si
2m
xample (2)
that
tane. + sece. Tane
sece
2m
sine
can30° - OM
(3) The angle! 45.
Consider the isosceles
triangle POR Vith
"PRO-900 and PR
____POR = / QPR =
PR
foras! theorem
RO
PQWZ
하하
tan45)
The angle 90-
- Consider the triangle
POR right angled at R
and £ POR = 90 where
e is small.
As
approaches zero,
QPR approaches 900
and length of PQ
approaches that of OR while PR approaches,
zerok
tane
se ine
sinoPR =
sing0° = =
Cose
PR
cos OPR
sine
CO'Se
1) co se
(Multiplying
Co se
Co se
Co
both the
umerator
tanQPR => OR
tangos
لم
Exerci
(1) Suppose
value of Lan
(2) Given
2mn
Find the
prove that
CO.SA
(ANES
sina:
sinB
(3) Suppose sine - msing and cose - ncos *
find the values of tane and tanx în term of m and n.
(4) Prove the following identities:
costand sind.
cosec.x - iseca
seca coseco-
sin+cot cosa
tan &
sind+ 2sino có sở
+ COS2+ cos
Cosecα
sin
sin
coseca + cot∞
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