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246. NOTE. In the next chapter we shall prove, that given three of the parts of a triangle (one of which must be a side), we can find the remaining three parts by means of the formulæ of this chapter. Hence, these formulæ cannot be equivalent to more than three independent equations. It is an instructive, but somewhat difficult problem to take three of these formula (e.g. A +B+C = 180°, b=c cos A +a cos C, c=bcos A +a cos B) and deduce all the others from them.

247. The following examples are important.
Example 1. Suppose we are given that

a=b cos C+c cos B,
b=c cos A +a cos C,

c=a cos B+b cos A. Then, taking a times the first +b times the second -C times the third, we get a+62 – =(ab cos C+ ac cos B) + (bc cos A +ba cos C)

- (ca cos B+ cb cos A), =2ab cos C.

a

sin ( [=d],

Example 2. Suppose we are given that A+B+C=180°,

b and that

sin A sin B sinĆ then 62 +62 - a do sino B+d2 sino C-d2 sin? A

2d2 sin B sinc sino B+sinoC-sin? A 2 cos A. sin B.sin C 2 sin B sinc

2 sin B. sinc [Ex, 29, p. 194, since (A+B+C)=180°.] =cOS A.

2bc

a

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sin (=d

248. The formula

b' sin A

sin B is very frequently of use in solving examples. Example. Prove that

a cos A +b cos B+c cos C=2a sin B sin C. Since a=d sin A, b=d sin B, crd sin C, the above may be written d sin A.cos A+ d sin B.cos B+ d sin C. cos C

=2d sin A .sin B. sin C, sin 24+sin 2B+sin 2C=4 sin A. sin B.sin C, which is Example (25) on page 194.

or

249. The student is advised to make himself thoroughly familiar with the following formulæ: a = b cos C + c cos B

(ii), 5

abc [= d]

(iii), sin A sin B

sin C

2S 78 +0% -as

(v). 2bc

a

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COS A =

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EXAMPLES. LXIV. In any triangle ABC

prove the following statements : sin A + 2 sin B sin C (1)

a + 20

C

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(3) a cos A +6 cos B-c cos C=2c cos A. cos B.

C A-B
(4) (a+b) sin
)

= CCOS
2

2
(5) (6-c) cos=a sin B-C

2
cos A

cos B (6)

+ sin B. sin c

= 2. sin C. sin À sin A. sin B

cos C

+

a-6

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с

a

(7) a sin (B-C)+b sin (C-A)+csin (A - B)=0.

cos B-cos A (8)

1+cos C btc cos B+cos C (9)

1-cos A

62 sin C+02 sin B (10) Wbc sin B.sin C

b+c (11) a+b+c=(6+c) cos A+(c+a) cos B+(a+b) cos C. (12) b+c-a=(b+c) cos A-(-a) cos B+(a - b) cos C.

a sin C (13) tan A

- acos

cOS A

cos B

+

a

C

(14) a (b + c) cos A +6(%+a%) cos B+c(a+b) cos C=3abc. (15) a cos(A+B+C) - cos (B+A)-C Cos (4+0)=0.

cos C a2 + 32 +% (16)

+ 6

2abc tan B a? +62 - c2

C

B (17)

(18) b cosa+c cosa tan C a? - 72 +20

B C 6+c-a (19) tan

tan

2. 2 b+c+ão (20) tan (6+c-a)=tan(c+a-6).

2

C (21) c2= (a + b)2 sina + (a - b)2 cosa

2

)

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** MISCELLANEOUS EXAMPLES. LXV. (1) If p is the length of the perpendicular from A on BC,

sin A=

ap

bc

(2) If 2 cos B .sin C= sin A, then B=C.

(3) If A=3B, then sin B=}

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(6) Given cos A=} and cos B=1, prove that cos C=-15. (7) If sin? B + sino C=sino A, then A=90°. (8) If sin 2B+sin 2C=sin 2A, then either B=90° or C=900.

(9) If A :B:C=1:2: 5, then 1+4 cos A. cos B.cos C=0, and a, b, c are in A.P.

A B-C B C-A (10) a sin

+ b sin

sin
2
2
2 2

A - B
+csin

= 0.

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z • sin

За

(11) If D is the middle point of BC, prove that

4AD2=262 +2c - a?. (12) Given that a= 2b, and that A=3B, prove that C=60o. (13) abc (a cos A +b cos B+c cos C)=8S2.

C B (14) If b cosa + cos2 then b, a, c are in A.P.

2

2 2 (15) If D, E, F are the middle points of the sides BC, CA, AB, then 4(ADP+BE+CF2)=3(a +62 +2).

22 - C2 (16) If D is the middle point of BC, cot ADB=

4S

(17) If d, e, f are the perpendiculars from A, B, C on the opposite sides of the triangle, then

a sin A +b sin B+c sin C=2(d cos Atecos B+fcos C). [Some of the Examples in the Appendix might be worked by the student at this stage.]

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