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237. II. To prove a= b cos C + c cos B.

From A, any one of the angular points, draw AD perpendicular to BC, or to BC produced if necessary.

There will be three cases. Fig. i. when both B and C are acute angles ; Fig. ii. when one of them (B) is obtuse; Fig. iii, when one of them (B) is a right angle. Then,

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a

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B C

B
CD
Fig. i.

=cos ACD; or, CD=b cos C,
CA

DB
and = cos ABD; or, DB = c cos B,

AB
...a=CD + DB=b cos C +ccos B.

CD
Fig. ii.

cos ACD; or, CD=b cos C,
CA
BD

= cos ABD; or, BD=c cos (180° – B),

AB
.: Q=CD-BD=b cos C - c cos (180° – B)

= b cos C + c cos B. Fig. iii. a=CB=b cos C

=b cos C + c cos B. [For, cos B= cos 90o = 0.] Similarly it may be proved that,

b=c cos A + a cos C; c=a cos B + b cos A.

238. III. To prove that in any triangle, the sides are proportional to the sines of the opposite angles; or, To prove

6 that sin A

sin C

a

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sin B

From A, any one of the angular points, draw AD perpendicular to BC, or to BC produced if necessary, Then,

AD
Fig. i. AD=b sin C; for, =sin C (Def.]:

AC

AD
also AD= c sin B; for, sin B.

AB
.:. b sin C=csin B;

6 or,

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sin B

sin c

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a sin A

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Q.E.D.

sin B

sin C

239. IV. Το

prove

that a= 5+ Co – 2bc cos A. Take one of the angles A. Then of the other two, one must be acute. Let B be an acute angle. From C draw CF perpendicular to BA, or to BA produced if necessary.

There will be three figures according as A is less, greater than, or equal to a right angle. Then,

Fig.I.

Fig.II.

Fig.III.

6

AZ 2

B

F

F
C
A B
A F B

A Fig. i. BC=C4 + AB - 2. BA.FA; [Euc. II. 13] or, ao = +C – 2c. FA

= 78 + c* – 2cb cos A. [For FA =b.coś A.] Fig. ii. BC=CA + AB+2.BA. AF; [Euc. 11. 12] or, a* = 52 + + 2cb cos FAC

= 32 + c 2bc cos A. [For FAC = 180o – A.] Fig. iii. BC=CA + AB;

[Euc. I. 47] or, a* = 68 + * — 2bc cos A. [For cos A = cos 90o = 0.] Similarly it may be proved that

7= c + ao – Aca cos B, and that

c* = a + 5* - 2ab cos C.

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2bc (6% + * — a®)_ a® – (* 2bc + c*) 2bc

26c a'-(6-c)'_ {1-(6-c)}{a+(6 – c)} 2bc

2bc A_ (a+c-b)(a + bc)

Q.E.D. 2

4bc

.. sin?

.

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А

+ -a
.:. 2 cosa 1 + cos A =1+

26c
A _ (b + c)' - a' _ (b + c + a)(b +c-a)
.: cos?
2 46c

4bc

Q.E.D.

242. Similarly it may be proved that

B (6+c-a) (b + a-c) B (a + b + c)(a + C-6) sin?

cos 2

2

4ca

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4ca

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so that

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a + b + c 243. VII. Now let s stand for

2 (a + b + c)= 28. Then, (6+c-a)=(6+c+Q – 2a) = (28 – 2a) = 2 (8a), and (c + a - b) = (c+a+b-26) = (28-26) = 2 (8-6), and (a + b - c)=(a+b+c-2c) =(28 - 2c) = 2 (8-c).

Then the result of Art. 241 may be written
A 2 (8-6) 2 (8-)

А (8-6) (8-C sin? 2 46c

2

bc

; or, sin

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А and cos

2 and so on.

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II

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2

COS

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Le Vo (8–a)(sB)(8–c).
The letter S usually stands for 18 (8 a) (8 - 6) (8-c), so

sin A 28 that the above may be written

abc

a

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