1969-12-02 — Page 14

高二第張四第一日三廿月十年酉巴瓣夏 WAH KIU YAT PO

郭日僑華

二期星日二月二十年九六九一藤公年八十五國民華中古教传

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1770英文中學會考試題預習專欄

堅道英文書院主編

數學科

(五)

MATHEMATICS (5)

Properties or a Triangle.

The tine zule To prove that

in any triangle. ABC.

sin B

Solution of any Triangle ABC. An element of a triangle

means either a side or an angle of the triangle, and

thus there are six elements to any triangle,

To solve a given triangle meanas given th independent elements to find the other three.

Note: The three angles of a triangle are not

independent since their sum is equal to two right angles.

Case (1) Three sides given, 1,07 2,

Kethod; Use the coming rule, i,e, COBA »

etc., to find A, B, C.

102

2bo

ample

Solve the triangle, a ches, o 4 inches,

6 inches

Case (1) Draw AD an altitude of the triangle Using Pythagoras Theorem,

2

BD

(AC2 - CD2) + BD2

AC2

CD2 + (BC — CD)2

OD?

LOE +

230.00 + 4"

28C. CD

Zab com 0

25

2bo

82.49

2-16--16

230

0.5652

C# C = -122 56 -0.75

2ab

41.25

Chaok, så + B + C - 82° 49'

jase (11) Two sides and included angle givan

'Usé by o and ▲)

othed

Use the cosine formula.

ocs à in order to find a' and then the rule to find B and C

2bo

cos C

-2abcost

Zab

Case (11) 10 – obtuse

Draw the perpendicular from a on 30 produced Using Pythagoras

2

* AD + BD2

CD2) + BD2

sxample 2.

SOLVE the triangle in whic b

321 in

o - 123 in, LA 29°16′

260 009

(3,21)2 + 2.2.in.

(1.23)2

swing the size rule,

sin

2(1.23)(3.21)(0.8723)

AC2 - CD2 + (BC + CD)< 103

Caso (1) Diagram (1)

Draw AD and BF the altitudes of tziingle ABC,

The AD AB sin B – AQ sin

a sin Bb sih O

Case (111) / 0 = 90°

sin B

ALSO BE = BC sin

in C

AB Sin

c.sin a

sin &

ain C

Using Pythagoras;

+ 2BG.CD + CD

2a.b

22/1.com (180°)

COR O

Zab ̃oom: 90%. (son90°-0)

Zaboos

2ab

Thus in all cases,

02.

→ 2að CORD.

Rab

Case (11) Diagram (ii)

Draw AD the perpendicular on BC produced, and BE the perpendicular on 40 produced.

AD ∞ AB sin 3 - AC sin (180°-C).

sin 3 – b sin (180o – 0) – b sin

sin B

The results for cos B and ces A can be proved år

• similar manner.

To prove that the area 4) or the triangle ABC 40" RIVER by A ̈*.bo sin A - ca sin 3 ■ub "in '0

Similarly, by using the altitude 33)

sin

sin

ein A

Bin

tase (iii) Diagram (iii) L© ■ 90°

10 - AB sin 3

Since sin 901

sin B

b sin 907

sin: 900

ein B

sin

Similarly

Hence, for any triangle ABC

sin A

"ne cosine rule

COBA M

200

Cos

Bin C

To prove that, in any triangle ABC,

20a

∙ase. (1) DE is an altitudo

Area of triangle ABC = CA.30 - 350

By drawing the other altitudes,

A- toa sin Bab sin G

A - Do sin a w♬ ca sin B = ƒ ab sin o

lase (11) Draw perpendicular BE from B on CA

produced

ACA, ĐÈ de Bin (180) A) thu B2.

By drawing in the other altitudes

▲ - I ca sia B - - eh sin G-

A-Do in Aca sin 3 = b ab sin

) (iii) A - 1 CẢ BA m the the nin 90

- 2 bó sin A

4180 ▲ - 1⁄2 bo - ✦ ba sin ✔ - on

Thus for any triangle ABC,

A-1 bo Bin A ✯ ca sin 3-« i ab sín

-0.7073

13:45 11 or 135 1!

As b is the longest side of the triangle

Therefore B should equal to 13591*

nuar

The solution is a - 2:2 in. /B

- 15.43

Cage (111) Given one side and two angles (6,

20

Since ▲ + 3 + 0 = 180", A. — 180′′ Thuˇsides "By o can now be found by using the sing

xample 3, Solve the triangle da which a × 10 inches,

£3 = 41°243 [0 = 35°18'

180° - 3-0

• 103°18"

180°- 41 24

5-18:

Using the MÄNG

Bin B

10 sin 41 241 Min 103°18

log þ➡ 1 + 10g(■in 41°24') - log(sin 103

+ 1.82041 - 1.98819 0.83222.

b. 6.796 incha.

win C

10 Bin 357187 Bin103°18′′.

log o = 1 + log(sin35°18')

10g(sin103 18

- 1 + 1.76182 - 1.98819 0.77363

5.938 inches

Example 4. In an obtuse-angled triangle ABC; the

angle B is 48 24, und the lengths of AB, 40 are respectively 5.38. and 4.29. Solve the triangle completely.

Using the sine rule,

sin

in C-

log sin

ain' 48"

10 - log5.38 + 1ogain48 24

073078 87378

10g 4.29 0.6324€

1.97210

or 180°

But the triangle is obtuse-angled;, thefore,

- 110 19 (A. cannot be obtuse sine / B

must at least be 69 41)

and

Hance / A = 180°

From the sine rule,

48°241 110°19' 21 17

b Bin A - 4:29.

sin 21 171 sin 48 24'

log a log 4.29 + log(sin 21°17,,A

ain B

· log sin (48°24')

0.63246 + 1.55989

1.87378 0.31857

a = 2.083

Therefore solution is, a

2.083,

110°191

- 21 17!