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VII.

SOLUTION OF PLANE TRIANGLES.

1. It is usually, though not always, best to work the propor tions in trigonometry by means of logarithms, taking tho Jogarithm of the first term from the sum of the logarithms of the second and third terms, to obtain the logarithm of the fourth term; or adding the arithmetical complement of the logarithm of the first term to the logarithms of the other two, to obtain that of the fourth.

2. There are three distinct cases in which separate rules are required.

CASE I.

3. When a side and an angle are two of the given parts, the solution may be effected by proposition 2 of the preceding section.

If a side be required, say,
As the sine of the given angle is to its opposite side,
So is the sine of either of the other angles to its opposite side.
4. If an angle be required, say,
As one of the given sides is to the sine of its opposite angle,
So is the other given side to the sine of its opposite angle.

The third angle becomes known by taking the sum of the two former from 180°.

Example 1.
Given angle A=24° 26'; angle
B=36° 43'; side b=137.6: to find

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Example 2. Given, sides a and b, as above, and angle A ; to find angle B.

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5. When two sides and the included angle are given, the solution may be effected by means of propositions 3 and 4. Thus, take the given angle from 180°; the remainder will be the sum of the other two angles.

Then, by proposition 3, -
As the sum of the given sides is to their difference,

So is the tangent of half the sum of the remaining angles to the tangent of half their difference.

Half the sum of the remaining angles added to half their difference will give the larger of them, and half their sum diminished by half their difference will give the lesser of them.

The solution may be completed either by proposition 4, or by proposition 2, as in Case I.

Example. Given side a=95.2, side b= 137.6, and the included angle = 118° 51'; to find the remaining angles. Here 180.00118° 51'=61° 09', the sum of the remaining angles.

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As sum of given sides, 232.8 .
Is to their difference, 42.4.
So is tany. I sum of rem. angles, 30° 34f'
To tang. their difference=6° 08}'

log. 2.366983 log. 1.627366 log. 9.771147 log. 9.031830

.

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Adding half the difference to half the sum, 300 341 +60 081=360 43', the larger angle, B. Deducting half the difference from half the sum=24° 26'= the smaller angle, A.

This case is susceptible of solution also by means of propo

CASE III.

6. When the three sides of a plane triangle are given, to find the angles.

First Method.. Assume the longest of the three sides as base; then say, conformably with proposition 5,

As the base is to the sum of the two other sides,

So is the difference of those sides to the difference of the segments of the base.

Half the base added to half the said difference gives the greater segment, and diminished by it gives the less; thus, by means of the perpendicular from the vertical angle, the original triangle is divided into two, each of which falls under the first

Or they may be solved by the simpler methods applicable to right-angled triangles.

case.

Second Method. 7. Find any one of the angles by means of proposition 6, and the remaining angles either by a repetition of the same rule, or by the relation of sides to the sines of their opposite angles.

VIII.

RIGHT-ANGLED PLANE TRIANGLES.

1. Right angles may be solved by the rules applicable to all plane triangles; and it will be found, since a right angle is always one of the data, that the rule usually becomes simplified in its application.

2. When two of the sides are given, the third may be found by means of the rule that the square of the hypothenuse is equal to the sum of the squares of the remaining sides.

3. Another method for solving right-angled triangles is as follows:

To find a side. Call any one of the sides radius, and write

Then say,

become sines, tangents, cosines, or the like, and write upon them the proper designations accordingly. Then say,

As the name of the given side is to the given side,
So is the name of the required side to the required side.

4. To find an angle. Assume one side to be radius, and mark the remaining sides as before.

As the side made radius is to radius,

So is the other given side to the name of that side;

Which determines the opposite angle.

5. Applying this method to the rightangled triangle A BC, and calling the hypothenuse a radius, we shall have,

c= a sin. C ; R; hence sin. C = Rc • a.
b= =a cos. C = R; hence cos. C= Rb • a.

Then, assuming the side b to be radius, we shall have,

c=b tang. C ; R; hence tang. C = Rc ; b.

If radius be called 1, the natural sines and cosines will be used in the application of these formulas; they are often more convenient than logarithms in railroad practice, especially when the numbers which measure the sides of the triangle are either less than 12, or are resolvable into factors less than 12.

6. The simpler relations between these natural functions are as follows:

sin.? + cos.? = 1; tang.

sin.

; cotang. COS.

COS. sin.

1 tamg.

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; versine=1-c08.; Co-versine=1-sin. sin.

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