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234. The following solutions may be applied to the third and fourth cases of oblique angled triangles; in one of which, two sides and the included angle are given, and in the other, the three sides. See pages 87 and 88.

CASE III.

In astronomical calculations, it is frequently the case, that two sides of a triangle are given by their logarithms. By the following proposition, the necessity of finding the corresponding natural numbers is avoided.

THEOREM A. In any plane triangle, of the two sides which include a given angle, the less is to the greater; as radius to the tangent of an angle greater than 45°:

And radius is to the tangent of the excess of this angle above 45°; as the tangent of half the sum of the opposite angles, to the tangent of half their difference.

In the triangle ABC, (Fig. 39.) let the sides AC and AB, and the angle A be given. Through A draw DH perpendicular to AC. Make AD and AF each equal to AC, and AH equal to AB. And let HG be perpendicular to a line drawn from C through F.

Then AC AB::R: tan ACH

And R tan(ACH-45°): : tan(ACB+B): tan(ACB-B)

Demonstration.

In the right angled triangle ACD, as the acute angles are subtended by the equal sides AC and AD, each is 45°. For the same reason, the acute angles in the triangle CAF are each 45°. Therefore, the angle DCF is a right angle, the angles GFH and GHF are each 45°, and the line GH is equal to GF and parallel to DC.

In the triangle ACH, if AC be radius, AH which is equal to AB will be the tangent of ACH. Therefore,

AC: AB::R: tan ACH.

In the triangle CGH, if CG be radius, GH which is equal to FG will be the tangent of HCG. Therefore,

R tan (ACH-45°):: CG: FG.

And, as GH and DC are parallel, (Euc. 2. 6.)

CG: FG::DH: FH.

But DH is, by construction, equal to the sum, and FH to the difference of AC and AB. And by theorem II, (Art. 144.) the sum of the sides is to their difference; as the tangent of half the sum of the opposite angles, to the tangent of half their difference. Therefore,

R: tan (ACH-45°)::tan (ACB+B) : tan (ACB~B)

Ex. In the triangle ABC, (Fig. 30.) given the angle A= 26° 14', the side AC=39, and the side AB=53.

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Tan 53° 39′ 9′′ 10.1332113 Tan (B– C)33° 8′ 50′′ 9.8149562

The same result is obtained here, as by theorem II, p. 75. To find the required side in this third case, by the theorems in section IV, it is necessary to find, in the first place, an angle opposite one of the given sides. But the required side may be obtained, in a different way, by the following proposition.

THEOREM B. In a plane triangle, twice the product of any two sides, is to the difference between the sum of the squares of those sides, and the square of the third side, as radius to the cosine of the angle included between the two sides.

In the triangle ABC, (Fig. 23.) whose sides are a, b, and c.

2bc: b2+c2 -a2::R: cos A

For in the right angled triangle ACD,
Multiplying by 2c,

But, by Euclid 13. 2,
Therefore,

bd::R: cos A 2bc: 2dc::R: cos A

2dc=b+ca-a3

2bc b2+c-a2::R: cos A.

The demonstration is the same, when the angle A is obtuse, as in the triangle ABC, (Fig. 24.) except that a2 is greater

than b+c; (Euc. 12. 2.) so that the cosine of A is negative. See art. 194.

From this theorem are derived expressions, both for the sides of a triangle, and for the cosines of the angles. Converting the last proportion into an equation, and proceeding in the same manner with the other sides and angles, we have the following expressions;

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These formulæ are useful, in many trigonometrical inveștigations; but are not well adapted to logarithmic computation.

CASE IV.

When the three sides of a triangle are given, the angles may be found, by either of the following theorems; in which a, b, and c are the sides, A, B, and C, the opposite angles, and h=half the sum of the sides.

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The quantities under the radical sign are the same in all

the equations.

In the triangle ACD, (Fig. 23.)

Rb::sin Ap. Therefore, sin Axb=Rxp.
√4b2c2 —(b2+c3 −a2)2. (Art. 221, p. 121.)

But p=

2c

This, by the reductions in page 122, becomes
√2h×2 (h− a) × 2 (h—b) × 2 (h—c)

p=

2c

Substituting this value of p, and reducing,

Sin A= =

2R

bc

√h (h—a) (h—b) (h—c)

The arithmetical calculations may be made, by adding the logarithms of the factors under the radical sign, dividing the sum by 2, and to the quotient, adding the logarithms of radius and 2, and the arithmetical complements of the logarithms of b and c. (Arts. 39, 47, 59.)

Ex. Given a=134, b=108, and c=80, to find A, B, and C. For the angle A.

h 161 log. 2.2068259
h. -a 27 log. 1.4313638
h-b 53 log. 1.7242759
h-c 81 log. 1.9084850

2)7.2709506
*3.6354753

log. 10.3010300

13.9365053

For the angle B.

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RX2

13.9365053

b 108 a. C. 7.9665762
C 80 a. c. 8.0969100

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By Art. 210, 2 Sin'A=R2-RXcos A. Substituting for cos A, its value, as given in page 132,

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This is the logarithm of the area of the triangle. (Art. 222.)

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But 2bc+a2 — b2 — c2 —a2 — (b −c)2=(a+b−c)(a−b+c) (Alg. 235.)

Putting then h= (a+b+c), reducing, and extracting ;

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Ex. Given a, b, and c, as before, to find A and B.

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By Art. 210, 2Cos2A=R2+Rxcos. A.

Substituting and reducing, as in the demonstration of the

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Putting h=(a+b+c), reducing and extracting,

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Ex. Given the sides 134, 108, 80; to find B and C,

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