(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 с 104. + + = 90°. 2 2 2 Trigonometrical Ratios are all positive.

and

the value of any one of its ratios is given. course apply to the angles B and C.

and

Now

EXAMPLE 1. To prove sin (A + B) = sin C.

EXAMPLE 2. To prove sin

4. tan A

A+B+C = 180°; .. A + B = 180° – C, ... sin (A + B) = sin (180° C) = sin C.

1. cos A = .

16.

A+ B+ C
2

..sin 4+ B

2

= - 1.

Therefore 4

A+ B
2

с

= COS

90°;

sin 90°

Find A from each of the six following equations, A being an angle of a

triangle:

15. tan A cot B cos C sec A csc B.

=

•

EXAMPLES. XXVI.

2. cos A =

5. √2 sin A = 1.

C

A+ B
2

A tan

is less than 90°, and its
A
Also, is known, when

2

Similar remarks of

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

8. cos (A + B + C ) = − 1.

7. sin (A + B + C ) = 0. 9. sin (A + B + C ) = 1. 11. tan (A + B) == tan C.

10. cos (A + B + C' ) = 0.
12. cot (B + C ) = tan † A.
14. cos (ABC) =

13. cos (A+B) = cos C.

= 90°

= COS

17.

sin 3 B sin 3 C
cos 3 C-cos 3 B

√3.20

105. 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,

A

A

b

A 4 4

b

a

D

C

B

B
II.
FIG. 50.

III.

Fig. i.

and

Fig. ii.

Fig. iii.

BD

AB

... a = CD – BD = b cos C - c cos (180° — B)

= bcos C+ e cos B.

= cos ABD; or, BD: = c cos (180° – B),

a CB b cos C

=

=

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

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

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

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

I. Fig. 50. AD = b sin C; for, = sin C [Def.];

AD
AC

or,

and

or,

=

=

...

AD = b sin C,

AD = c sin ABD = c sin (180° – B).

.. AD c sin B;

=

.. b sin Cc sin B;

C

sin C

Similarly it may be proved that

a

sin A

с

b
sin B sin C
III. Fig. 50. AB AC · sin C'; or, cb sin C;
b
sin B

=

=

sin B.

=

C

sin C

=

b sin B

a

b sin A sin B

―

107. IV. To prove that a2 = b2 + c2 - 2 bc 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.

a

A A A

b

В

A B C

F

III.

II.
FIG. 51.

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

or,

or,

or,

I. Fig. 51.

and that

cos A

=

108. V. Hence,

=

II. Fig. 51. BC2 = CA2 + AB2 + 2 · BA · AF;

a2 = b2 + c2 + 2 cb cos FAC

b2 + c2

EXERCISE I.

· b2 + c2

III. Fig. 51. BC2 = CA2 + AB2;

and

and

Similarly it may be proved that

b2 + c2 — a2

-

2 bc

cos B

2 cb cos A.

a2 = b2 + c2 2 be cos A. (For cos A=cos 90°=0.)

b2 = c2 + a2 − 2 ca cos B,

c2 = a2 + b2 - 2 ab cos C.

=

.. cos A

2 be cos A. (For FAC = 180° – A.)

(Geom.)

c2 + a2 62
2 ca

(For FA= bcos A.) (Geom.)

*109. The formulæ of Art. 108 may be obtained directly from those of Art. 105.

a = b cos C + c cos B.

(1)

b = ccos A+ a cos C.

(2)

c = a cos B+ b cos A.

(3)

Multiplying (1), (2), (3) by a, b, c respectively and adding, we

(Geom.)

Cos C

obtain

a2 + b2 + c2 = 2 a (b cos C+ccos B) + 2 be cos A = 2 a2 + 2 be cos A.

a2

=

cos C.

EXERCISE II. If a = 5, b = 6, c = 7, find cos A.

a2 + b2 — c2

2 ab

Find the two corresponding expressions, viz., for cos B and

110. VI. Let s stand for half the sum of a, b, c; so that (a + b + c)=2 s.

Then, (b+ca) = (b + c + a2a)=(2s-2 a)=2(s− a),

(c+ab)=(c+a+b-2b)=(2s-2b)=2(s—b),

(a + b − e)=(a+b+c−2 c)=(2 s− 2 c)=2(s — c).

111. VII. To prove that

(s — b) (s

bc

sin

112.

=

V

cos A

2 sin2 4

2

sin?4

2

2

where s stands for half the sum of the sides a, b, c.

Now, since

..sin:

.. sin

tan

=

=

=

2

22 = V

Again, since 2 cos24

1

b2 + c2 a2
2 bc

哈

=

2 bc − (b2 + c2 — a2
2 bc

sin

(2 s

A

2

A

COS
2

cos A 1

=

(a + c − b) (a + b −c);

4 bc

113. VIII. Again,

sin A

sin

c)

... sin A

a2 − (b − c)2 _ { a − (b − c)} {a + (b − c)}
2 bc
2 bc

=

- 2 b) (28 -
4 be

=1+ cos A,

=

1+ cos A=1+

28.

and that cos

and 1

.. sin A2.

2 sin

b2 + c2 — a2
2 bc

b2 + c2- a2
2 be
(b + c)2 — a2 _ (b + c + a) (b +
2 be

2 bc

=

2

Vs (s bc

- 2 c)

2 a)

cos A = 2

A

A

14. cos 4;

2

bc

=

a

b) (s — c)
bc

s (s
bc

EXAMPLE. Write down the corresponding formulæ for
B

B

B for cos 2

and for tan R22.

2

=

(s—b) (s—c)

V

a2 (b2 - 2 bc + c2)

2 bc

S (S

2 sin24 (Art. 88.)

s ( s − a).
be

a) (s — b) (s —

b) (s — c).

bc

S

bc

b)

C

s (s - a)

bc

a)