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8. Show that the ratios of any angle, whether acute or obtuse, are subject to the relations of equality established in Chapter III.

9. How must the relations of inequality of Chapter III. be modified in order to include obtuse angles ?

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RELATIONS BETWEEN THE SIDES AND THE RATIOS

OF THE ANGLES OF A TRIANGLE.

141. Consider any triangle ABC.

The angles at the points A, B, C will be called by the names A, B, C.

The sides BC, CA, AB, which are opposite to these angles, will be called a, b, c respectively.

142. To draw the figure by which the ratios of the base angles of a triangle may be indicated.

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Draw AL perpendicular to BC or BC produced. Then, the angle ACB may be (i) acute, (ii) obtuse, or (iii) right. In each case,

LA sin ACB

.. CA sin ACB=LA.

CA' But in case

CL (i) cos ACB=

.. CA cos ACB=CL. CA'

- CL (ii) cos ACB=

.. CA cos ACB =-CL. CA

0
(iii) COS ACB= .. CA cos ACB= 0.

CA'

143. To prove the double-cosine formula:

b cos C + c cos B=a.

BL In each fig.

.. BA cos B = BL. BA'

CL In fig. (1)

.. CA cos C = CL. CA'

COS B

cos C

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.: CA sin c BA sin B, i.e. b sin C=c sin B. COR. Dividing by bc, we have

sin C sin B

6

с

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sin B sin A Similarly

b

a
sin A sin B sin C

6 i.e. the sines of the angles are proportional to the opposite

a

с

a

с

145. By the corollary of the last article, we have

6 sin A

sin B sin C. Put each of these fractions equal to d. Then since in each fraction the numerator is a length and the denominator a ratio, therefore d will represent a length.

Also, since

a sin A

.. a= d sin A.

Similarly,

b=d sin B and c= d sin C.

=

146. To find the sine and cosine of the sum and difference of two angles by means of the double-cosine and double-sine formula. In the equation

a=b cos C + c cos B, substitute for a, b, c in terms of d.

[Art. 145. Thus d sin A = d sin B cos C + d sin C cos B,

.. sin A sin B cos C + cos B sin C. (A.)
Similarly sin B = sin C cos A + cos C sin A.
i.e. (substituting for sin A)

sin B=sin C cos A + cos2 C sin B + cos B cos C sin C,
.. - sin C cos A = cos B cos C sin C – sin B (1 - cosa C),
..
= cos B cos C - sin B sin C.

(B.) Now sin A sin (B+C) and - cos A = cos (B+C), [Art. 140. :. (A) and (B) become

sin (B+C) = sin B cos C + cos B sin C (1),

cos (B+C) = cos B cos C – sin B sin C ......... (2). Let B' be the exterior angle at B. Then A= B'-C, sin B=sin B' and cos B=- cos B', :: (A) and (B) become

sin (B' C) = sin B' cos C – cos B' sin C ..(3), cos (BC)=cos B'cos C + sin Bsin C .(4).

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147. It should be observed that, by the method of the last article, we have proved the formulæ for the sine and cosine of the sum and difference of two angles, whatever geometrical value the angles considered may have, i.e. whether they are acute or obtuse.

Hence, also, the formulæ which may be algebraically derived from these are universally true for all geometrical angles, viz. the expressions for the tangent and cotangent of the sum or difference of two angles, and the ratios of the double angles, &c. Example. If A, B, C are the angles of a triangle, express

sin A+sin Btsin C as a product of ratios.

The given expression
=sin {t (A+B)+} (A B)} +sin {I (A+B) }(A B)} +sin C
= 2 sin 3 (A+B) cos } (A - B)+2 sin cos C
=2 cos } C {cos } (A B)+cos } (A + B)},

[:: sin (4+B)=cos C and sin C=cos (A+B)]
=2 cos } C. 2 cos ; A cos } B
=4 cos } A cos B cos } C.
148. To prove the cotan-cosec formula :-

a

cot B + cot C= cosec B or

cosec C. C

7 CL

- CL In fig. (0)

cot C=
LA ; in fig. (ii) cot C=

LA

BL In each fig.

cot B =

LA Again, in fig. (°), BL+CL = BC;

in fig. (ii), BL-CL= BC. .. in each case,

BC BC AB cot B + cot C

LA - AB LA

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This formula may be written in other forms: thus

b sin C 6 tan C tan B = a-b cos C

a sec C -6°

149. To prove the cosine formula :

cm = a2 + 62 - 2ab cos C.
In fig. (i), (Euc. II. 13), ABP = BC + CA - 2BC.CL.
In fig. (ii), (Euc. II. 12), AB? = BC2 + CAP + 2BC.CL.
In fig. (iii), (Euc. I. 47), ABP = BC2+CA?
But in fig. ()

CL=CA cos C,
in fig. (ii)

CL=-C A cos C, in fig. (iii)

0=CA cos C. i. in each case,

ABP = BC2 + CA 2BC.CA.cos C, i.e.

C = a + b2 - 2ab.cos C. Similarly

62 = c + a2 - 2ca.cos B, and

a = b2 + - 2bc.cos A.

150. To prove the sine formula :

2ab sin C = {{(a+b+c) (a + bc) (c+a b)(b +c-a)}. By the last article,

2ab cos C = a2 + 62 - c*, .. 2ab (1 + cos C) = (a? + 2ab +62) – co = (a + b + c)(a + b – c) (1),

2ab (1 - cos C) = c – (ao – 2ab + b*) =(c + a-6) (c— a+6) (2). .. multiplying the above equations (1) and (2), we have 4a2b2 (1 – cosC) = (a + b + c) (a + b c) (c+a6).(b+c-a); .. taking the square root, since 1-cos C = sino C,

2ab sin C= _{(a+b+c) (a + b -c) (c+a-6) (6+c-a)}. 151. To prove the half-angle formule.

Since 1 + cos C = 2 cosa fC and 1 - cos C = 2 sinC, substituting in (1) and (2) of last article

4ab cosa 4C = (a + b + c)(a + b -c),

4ab sino 4C = (c+a-6)(c - a+b). Dividing, we have

(c+a-6) (6 + c-a) tano 4C =

(a + b + c)(a + bc)

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