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 formula 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=b cos A+ a cos B) and deduce all the others from them.

A

tan

Then, taking a times the first +b times the second the third, we get

a2+b2 - c2=(ab cos C+ ac cos B) + (bc cos A+ba cos C')

=2ab cos C.

-c times

-(ca cos B+cb cos A),

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

sin2B+sin2C-sin2 A 2 cos A. sin B. sin C

or

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.

249. The student is advised to make himself thoroughly familiar with the following formulæ :

(3) a cos 4+b cos B- -c cos C=2c cos A. cos B.

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

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

(14) a (b+c) cos A+b (c+a2) cos B+c (a2+63) cos C-3abc. (15) a cos (A+B+C)−b cos (B+A) − c cos (4+C)=0.

99

(16)

COS A
+

α

b

COS B
+

с

a2+b2 - c2

(17)

@

(19) tan

B

2

14

2abc

C

(18) b cos2+c cos2:

(20) tan (b+c-a)-tan (e+a-8).

(21) c2=(a+b)2 sin2 + (a - b)2 cos2

B

=S.

** MISCELLANEOUS EXAMPLES. LXV.

(1) If p is the length of the perpendicular from A on BC,

(6) Given cos A= and cos B=13, prove that cos C= -1.

(7) If sin2 B+ sin2 C=sin2 A, then A=90o.

(8) If sin 2B+sin 2C=sin 24, then either B=90° or C'=90°.

(9) If A: B: C=1: 2: 5, then 1+4 cos A. cos B. cos C=0, and a2, 62, c2 are in A.P.

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

4AD2=262+2c2- a2.

(12) Given that a=26, and that A=3B, prove that C=60". (13) abc (a cos A+ b cos B+c cos C)=8S2.

If D, E, F are the middle points of the sides BC, CA,

(15)

AB, then

(16)

b2-c2

4.S

4(AD2+BE+CF2)=3(a2+b2+c2).

If D is the middle point of BC, cot ADB=

(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+csin C-2 (d cos A+e cos B+fcos C).

[Some of the Examples in the Appendix might be worked by the student at this stage.]