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

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

and

Now

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

A+B+C 180°; .. A + B = 180° — C,

=

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and

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2

EXAMPLES. XXVI.

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

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

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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,

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=

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b cos C+ c cos B. [For, cos B :

Similarly it may be proved that,

= cos 90° = 0.]

bc cos 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

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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

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Take one of the angles A. acute.

sin A

sin B

a

sin A

=

b

sin B

=

с

sin C

a2 = b2 + c2 · 2 bc cos A.

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Q.E.D.

Then of the other two, one must be Let B be an acute angle. From C draw CF perpendicular to BA, or to BA produced if necessary.

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There will be three figures according as 4 is less, greater than,

or equal to a right angle. Then,

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or,

or,

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CA2+ AB2-2. BA FA;

a2 = b2 + c2 2c. FA

(Geom.)

= b2+c2-2 cb cos A. (For FA= bcos A.)

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

a2 = b2 + c2 + 2 cb cos FAC

(Geom.)

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**

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

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

obtain

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

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EXERCISE I.

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

cos C. EXERCISE II.

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

and

110. VI. Let s stand for half the sum of a, b, c; so that

(a+b+c)=2s.

Then, (b+c − a) = (b+c+a-2a)=(28-2a)=2(sa),

(c+ab)=(c+a+b2b) = (2 s−2b)=2(s — b),

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

bc

a),

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