F-DHAB BATAOKE WAH KIU YAT PO
英中會考附加數學(二)答案
堅道英文書院答
SOLUTIONS TO HONG KONG
CERTIFICATE OF EDUCATION EXAMINATION)
(ENGLISH) 1970
ADDITIONAL MATHEMATICS PAPER II
SECTION B
10.
OB & 500 30 OC OB sec 30°
OD * 00 sec 30°
(ii) AB
вес
報日僑華
三期星
日十二月五年〇七九一曆公年九十五國民華中
育教僑華
tan 34 cot 2A becomes
글
•+3)
* (1
+2)(1
5+4 10+2
Since A 18°is a root of tan 34 - cot 24,
it is also the root of 5tan A-
or tan 18 is the root of 5x The root of the last equation
• [ 10 € (102 - 4 x 5.]
+
2. x. 5
10tan 2
(15) From the figure 180
8 be the mid-pt. of PQ
SQ2
DC
2 DS
90°
Since tan 18 is greater than zero,the positive sign is used outside the bracket And also tan 18° is smaller than one, the négative sign is used inside the bracket. Answeri tan 18 - (1
Take as the origin of a co-ordinave syɛTOM, Positive sign is taken in the direction upward and to the left. With respect to thiB co-ordinate system
■ (1 + cos(180 -/3), 4 + Bin3
Bina, I
QUB :)
sec 30°
sec 30
(1) 0,
fore,
181
(tan 30°) a
OB tan 30°
a sec 30 tan 30°
(3 + ain/b + cosa
14 = 20 + 12 cos
Dos
120
when x = 1 or 2 and y
4 resp.
A-90-30%
(iii) length BCDE 18
30+ 120 180°
2.617 +3.74
6,4 miles (2 #18.-11gure}
CD • OC tan
& sec 30)
tan 30%
- OD tan 30°
a sec
tan 30.
when
AB:BC:CD DE 1 1 dec 30
sec2 30° 180c 30°
That is the four lengthe form a geometrical progression with the common ratio seo 30°,
when
(iii) Perimeter of the figure OABCDE IB
tan 30° + e tan30 sec 30°
atan 30 seo 30° a sec 30:
(22 +375)
6(2x = 3)
d2y/4'x is less than 0, (1,5) is a maxiumu point.
y is greater than 0, (2,4) is a minimum point.
-2(x+2)
when 1 - -2 and y
2 which is greater than 0
(~2,-4) is a minimum point.
a( 32+ 712)
(11) The x coordinate of the intersecti
are given by the roots of
pointm
Since tax
90 >e >
(39 + 7/3)
2x
12x
and Otan
100; 100 € 45%
2x - 10x
Area of OAB - OA.AB-
Area of A OBC - 2 08.BC-
Area of ▲ och = 1⁄2 OC.CD -
Area of ▲ ODE = 1 OD. DE
atan 30°
2x(x −1) (x
tan 30 sec. 30
0,5, 32
espectively
tane, tane,
tan
30° Bac+30°
Equation of tangent to 2 at (0,0).
tan 30 sec 30o
gradient at (0,0)
(0 + 4).
The areas above are in in geometrical progression with the common ratio of ̃*•
(v) Area of DABCDE
tan 30°(1 + sec2 30° + sec130°
175122)
+ ₤ +(4/3)2 + (4/3)?
10. 30-
,(1,5).
(4,32)
12
12x-16
|(8,0)
tan 2/ AXD
John (8, 8) to the origin and find
tane
ninos
is a possible gradient for
(8, 8) is a possible point for F.]
(iii) Using (ii) we know the circle must pass the
three points (0, 0), (8, 0), (8, 8)
The centre of the circle is (4,4) and the
Tadius is 1/8282
4J2
tan 7IXE
4/3.
tan
_ AYB - 2 x —
angles AXB and AYB together egual
90 degrees.
J4;
the formula tan(A+B)
tan 3A cot 24.
tan Atan B
* tanà tán B.
let A - 18o, tan 34 - tan 54°
cot 24 cot 36°.
therefore 18 is a root of the equation.
tan Atan A
Tan 34
tan
cot 2A = 1/tan2A
2tan A
Let t tan A
fis.
Equation of aircle
(iv) At the points of intersection of the circle
and y-.
wo have
5x2 - 32x
(x-8)
(- Kx-8) - 4]
B
From the diagram PR
(1)
2
PR
18
18 - 3/2 miles
O respectively
The mid point is (pq) – (b[8 - 3), 1⁄2 24)
pzq-16 12
(未完轉入第六張第二百