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CHAPTER XVI.

ON THE RELATIONS BETWEEN THE SIDES AND ANGLES

OF A TRIANGLE.

231. The three sides and the three angles of any triangle, are called its six parts.

By the letters A, B, C we shall indicate geometrically, the three angular points of the triangle ABC ; algebraically, the three angles at those angular points respectively.

А

c

a

с

B By the letters a, b, c we shall indicate the measures of the sides BC, CA, AB opposite the angles A, B, C respectively.

232. I. We know that, A + B+C = 180°. [Euc. 1. 32.]

233. Also if A be an angle of a triangle, then A may have any value between 0° and 180°. Hence,

(i) sin A must be positive (and less than 1),

(ii) cos A may be positive or negative (but must be numerically less than 1),

(iii) tan A may have any value whatever, positive or negative.

234. Also, if we are given the value of

(i) sin A, there are two angles, each less than 180", which have the given positive value for their sine.

(ii) cos A, or (iii) tan A, then there is only one value of A, which value can be found from the Tables.

+

A B

А 235. +

= 90°. Therefore is less than 90°, 2 2 2

2

A and its Trigonometrical Ratios are all positive. Also, is

2 known, when the value of any one of its Ratios is given. Similar remarks of course apply to the angles B and C. Example 1. To prove sin (A+B)=sin C.

A+B+C=180° .: A+B=180° C, and .. sin (A+B)=sin (180° C)=sin C.

[p. 104.] A+B С Example 2. To prove sin

2

=COS

2

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Find A from each of the six following equations, A being an angle of a triangle. (1) cos A=1 (2) cos A -7. (3) sin A=1.

1 (4) tan A=-1. (5) sin A

(6) tan A -3.

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Prove the following statements, A, B, C being the angles of a triangle.

sin (A+B+C)=0. (8) cos (1+B+C)= -1. (9) A+B+C

A+B+C
sin

=l.
(10) cos

=0.
2

2

B+C А (11) tan(A + B)= -tan C. (12) cot tan

2

2 (13) cos (A+B)= - cos C. (14) cos(A+B-C)= -cos 2C. (15) tan A - cot B=cos C. sec A . cosec B.

sin A - sin B c A-B (16)

=tan tan sin A+ sin B 2

2 sin 3B -sin 3C 3A (17)

tan cos 3C - cos 3B 2

A B

sin

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COS

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A B с

A B C
-sina = +sin? 1-2 cos

sin COS
2
2

2 2 2
B+C-A C+4 B A+B-C
(24) sin

+sin
+sin

1
2
2

2

А A B
=4 sin sin sin

2 2
(25) sin 2 A+sin 2B+sin 2C=4 sin A . sin B. sin C.
(26) sin A.cos A - sin B. cos B+sin C. cos C

= 2 cos A. sin B. cos C. (27) sin(B+C-A) -sin (C+A - B)+sin (A+B-C)

=4 cos A. sin B. cos C. (28) cos 2A +cos 2B+cos 2C=-1-4 cos A. cos B. cos C. (29) sin? A - sino B+sino C=2 sin A . cos B. sin C. (30) cos (B+C - A)+cos (C+A - B) cos (A+B-C)+1

=4 sin A.sinB . cos C. B C B C А (31) sin

+ sin
2 2 2 2 2 2

А B A B с
+sin
COS COS

=sin

sin
2 2 2 2 2

ž (32) tan A +tan B+tan (=tan A . tan B. tan C.

B c c A А B (33) tan . tan + tan tan

tan -1.

2

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COS

COS

COS

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236. If ABC be a right-angled triangle having C = 90°, then (A + B) = 90°.

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a

In a right-angled triangle ABC, in which C is a right angle, prove the following statements. (1) tan A=cot B.

(2) tan B=cot A +cos C. (3) sin 2A=sin 2B.

(4) cos 2A +cos 2B-=0. 2ab

6 (5) sin 2A c. (6) cosec 2B= +

26

2a. 72 - a?

sin? A - sin? B (7) cos 2 A

(8) cos 2B=

sin? A+sino B B (9) sin?

A c+b 2

(10) cos?
2c

2 2c
A
atc

A - B (11) COS + sin

(12) =tan 2 2

atb

2 (13) sin(A - B)+cos 2A=0. (14) sin (A - B)+sin (2A+C)=0. (15) (sin 4-sin B)2+(cos A +cos B)2=2.

C-a

a-6

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a+ (1

2 sin A + - 6 a+ bcos 2B

la-b

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