1970-05-20 — Page 22

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