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RULE.

1. To find the other opposite side.
Sine of the angle opposite to the given side,
Is to sine of the given side;
As the sine of the other given angle,

Is to the sine of its opposite side. To the side found by this proportion, and its supplement, add the given side. Then if each of these sums be of the same species with respect to 180°, as the sum of the given angles, the problem is ambiguous; that is, the side thus found may be either acute or obtuse.

But, if only one of these sums be of the same species as the sum of the given angles, that value of the side, found by this proportion, must be taken, which when added to the given side agrees with the sum of the angles. In this case the problem is not ambiguous.

2. To find the side adjacent to the two given angles. Find the side opposite to the other given angle, by the first part of the rule, and note whether it be acute, obtuse, or ambiguous. Then, Sine of half the difference between the two given

angles,
Is to sine of half their sum;
Aktangent of half the difference between the two

sides,
Is to tangent of half the third side. (N. 189.)

3. To find the third angle. Find the side opposite to the other given angle, by the first part of the rule, and note whether it be acute, obtuse, or ambiguous. Then, Sine of half the difference between the two sides

containing the required angle, Is to sine of half their sum; As tangent of half the difference between the other

two angles, Is to cotangent of half the required angle. (M.188.)

CASE III. (N) When two sides and the included angle, of an obliqueangled spherical triangle, are given, to find the rest.

RULE.

1. To find the other two angles.
Cosine of half the sum of the two given sides,
Is to cosine of half their difference;
As cotangent of half the included angle,

Is to tangent of half the sum of the other two angles. Half the sum of these two angles must be of the same species as half the sum of the given sides.

Secondly,
Sine of half the sum of the two given sides,
Is to sine of half their difference;
As cotangent of half the included angle,
Is to tangent of half the difference between the other

two angles. (M. 188.)
Half the difference between these angles is always acute.

Lastly, Half the sum of the two angles increased by half their difference, gives the angle opposite to the greater side, and diminished by the same, leaves the angle opposite to the less side. (C. 35.)

2. To find the third side. Find the two required angles by the first part of the rule. Then, Sine of half the difference between these angles,

Is to sine of half their sum;
As tangent of half the difference between the given

sides,
Is to tangent of half the third side. (N. 189.)

OR, without finding the other two angles. To the sum of the logarithmical sines of the given sides, add double the logarithmical sine of half the contained angle, and reject 30 from the index.

Look for the remainder in the table of logarithmical sines, and take the degrees and minutes answering

to it. Then take the difference between twice the natural sine of those degrees, and the natural cosine of the difference between the given sides; the remainder will be the natural cosine of the side required. This side is acute or obtuse, according as the double natural sine is less, or greater, than the natural cosine of the difference between the given sides. (X. 192.)

CASE IV. (O) When two angles of an oblique-angled spherical triangle, and the side adjacent to both of them, are given to find the rest.

RULE.

1. To find the other two sides.
Cosine of half the sum of the two given angles,
Is to cosine of half their difference;
As tangent of half the adjacent side,

Is to tangent of half the sum of the other two sides. Half the sum of these sides, must be of the same species as half the sum of the given angles.

Secondly.
Sine of half the sum of the two given angles,
Is to sine of half their difference;
As tangent of half the adjacent side,

Is to tangent of half the difference between the other two sides.

Half the difference between these sides is always acute. (N. 189)

Lastly. Half the sum of the two sides increased by half their difference, gives the side opposite to the greater angle, and diminished by the same, leaves the side opposite to the less. (C. 35.)

2. To find the third angle. Find the two required sides by the first part of the rule. Then, Sine of half the difference between these sides,

Is to sine of half their sum;
As tangent of half the difference between the given

angles,
Is to cotangent of half the third angle. (M. 188.)

Or, without finding the other two sides. To the sum of the logarithmical sines of the given angles, add double the logarithmical cosine of half the given side, and reject 30 from the index.

Look for the remainder in the table of logarithmical sines, and take the degrees and minutes answering to it. Then take the difference between twice the natural sine of those degrees, &c. and the natural cosine of the difference between the given

angles; the remainder will be the natural cosine of the angle required. This angle is acute or obtuse, according as the double natural sine is greater, or less, than the cosine of the difference between the given angles. (Y. 193.)

CASE V. (P) When the three sides, of an oblique-angled spherical triangle, are given to find the angles.

RULE I.

From half the sum of the three sides subtract the side opposite to the required angle, and note the half sum and remainder. Then add together,

The logarithmical co-secants of each of the sides containing the required angle, rejecting the indices; and the sines of the above half sum and remainder: half the sum of these four logarithms is the logarithmical cosine of half the angle sought. (G. 185.)

or, RULE II. Add all the three sides together, from the half sum subtract each side containing the required angle, and note the remainders. Then add together,

The logarithmical co-secants of each of the sides containing the required angle, rejecting the indices; and the sines of the above-noted remainders: half the sum of these four logarithms, is the logarithmical sine of half the angle sought. (F. 184.

OR, RULE III.
From half the sum of the three sides subtract each side

separately. Then add together,

The logarithmical co-secants of half the sum of the sides, and of the difference between that half sum and the side opposite to the angle required, rejecting the indices; the logarithmical sines of the difference between the half

sum and each side containing the required angle, half the sum of these four logarithms is the logarithmical tangent of half the angle sought. (H. 186.)

CASE VI. (Q) When the three angles, of an oblique-angled spherical triangle, are given to find the sides.

RULE I.

Add all the three angles together, take the difference between the half sum and the angle opposite to the side sought, and note the half sum and remainder. Then add together,

The logarithmical co-secants of each of the angles adjacent to the required side, rejecting the indices, and the cosines of the above half sum and remainder; half the sum of these four logarithms is the logarithmical sine of half the side sought. (I. 186.)

OR, RULE II. Take the supplements of each of the angles, and use the remainders as sides in a new triangle.

Find the angles of this triangle, by any of the rules in Case V. the supplements of which will be the sides sought. (U. 137.)

(R) Case I. Given two sides of an oblique spherical triangle, and an angle opposite to one of them, to find the rest.

In the oblique spherical triangle ABC.

The side ac=80°.19'
Given The side bc=639.50 Required the rest.

The LA=519.30

BY CONSTRUCTION. (Plate V. Fig. 15.) 1. With the chord of 60 degrees describe the primitive circle; through the centre p draw cpe, and apr at right angles to it.

2. Set off the side ac —80°.19' from c to A, by the scale of chords.

3. Through a draw the great circle abon, making an angle of 510.30' with the primitive. (P. 160.)

4. Set off the side bc=639.50' by a scale of chords, from c to m, and draw the parallel circle mbem. (Z. 162.) Through the points b, B, where it cuts the oblique circle abon, and the point c, draw the great circles cbe, CBe.

5. Then, abc or ABC is the triangle required, each having the same data, which shews this example to be ambiguous.

To measure the required parts. 6. The side Ab (C. 163.)=280.33', and AB=120°.47'. 7. The Lach (G. 164.)=240.37', and L ACB=131o.29';

the 2 abc=1200.44', and the L ABC= 59o.16'.

sine BC=

=630.50 : sine LA=510.30' :: sine ac=80°,19' : sine , B=59o.16'

BY CALCULATION.
1. To find the LB.
9.95304 Because AC + BC, A + (B acute), and a to
9.89354 (B obtuse) are each of the same species

9.99377 with respect to 180°, the LB is ambigu: 9.93427 ous (Y. 232.) being=59o.16' or its sup

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