XI. Given the three sides to find either of the angles.

RULE.

As rect. under the sine of the sum of the three sides and the sine of the difference between this sum and the side opp. the sought square of rad.

So is rect. under the sines of the differences of the samesum and each of the other two sides: square tan required 4.

Or,

If s denote the sum of the three sides, c the side opp. the required, and a, b the sides about that 4.

Ltanc

sins sin (s- -c)

Lsins+€Lsin (1s−c) +Lsin (1s−a) +L sin (↓ s—b)

2

Which is always acute.

Or this logarithmic formula may be expressed in words, as follows:

Add together the log sine of the sum of the three sides, and the log sine of the difference between this sum and the side opp. the angle sought, and find the complement of their sum, by taking the index from 19, and the rest of the figures from 9, as usual.

To this complement add the log sines of the differences between the same sum and each of the other two sides, and the result, divided by 2, will give the log tan ofthe required 4.

Or the whole angle may be found by the following formula :

Note. In this case the three sides must be so taken, that the sum of any two of them may be greater than the third, and the sum of all three of them less than 360°.

XII. Given the three angles, to find either of the sides.

RULE.

As rect. under the cosine of the sum of the three

S

Zs and the cosine of the diff. between this

sum and

the opp. the side sought

square of rad.

So is the rect. under the cosines of the differences

of the same

square of cot

sum and each of the other two 4s:

required side.

Or,

If s denote the sum of the three Zs, c the Z opp. the required side, and A, B the 4s adjt. to that side.

Cot & c=r√

1. cot &c=

EL COS 1S+EL COŚ ́1 S-C) +L COS(+S−A) +L COS(S-B)

2

Or this logarithmic formula may be expressed in words, as follows:

S

Add together the log cosine of the sum of the three and the log cosine of the diff. between this sum and the opp. the side sought, and find the complement of their sum, by taking the index from 19, and the rest of the figures from 9, as usual.

To this complement add the log cosines of the differences between the same sum and each of the other

twỏ ▲3, and the result, divided by 2, will give the log cot. of the required side.

Or the whole side may be found by the following formula:

Note. In this case the three angles must be so taken, that the sum of any two of them may be greater than the supplement of the third, and that the sum of all three of them may fall between 180° and 540°.

REMARK,

When the triangle, in any of the cases of the preceding table, is isosceles, it may be readily resolved by one or other of the four following analogies, which taken directly, or by permutation, contain all the varieties of this kind that can happen in practice.

1. As rad sin vert. :: sin either side: sin the base.

2. As rad: cos either at base :: tan either side : tan the base.

3. As rad tan either at base :: cos either side : cot vert. 4.

4. As rad sin either at base :: cos the base : cos vert. 4.

Where it is to be observed, that when the base, or the vert. becomes the 4th term of the proportion, or the thing sought, it is always less than 90°.

And that when one of the equal sides, or angles, is the thing sought, it will be like its opposite angle,

N

or its opposite side, according as it is a side or att angle (o).

Also, if the triangle have two of its sides, or two of its angles supplements of each other, it may be resolved by one or other of the four following analogies, which have the same generality as the former.

1. As rad cos vert.

the base.

:: sin either side: cos

2. As rad: cos either at base :: tan its opp. side cot the base.

3. As rad: tan either at base ::

4. As rad sin either 4 at base :: sin

; sinvert. 4.

cos its opp.

the base

In which cases, when the base, or the vert. 4, is the 4th term, or thing sought, it is always less than 90°.

And when a side, or an angle, is the thing sought, it will be like its opposite angle or opposite side, as in the isosceles triangle.

(0) As every isosceles spherical A is divided into two equal rightangled spherical As, by a perpendicular drawn from the vertical 4 to the base, it is evident that the solutions of all the cases of the former may be derived from those of the latter, by referring them to the right-angled A to which they belong.

Also, in the species of ▲ which has the sum of two of its sides equal to 180°, the sum of their opposite will likewise be equal to 180°; and it is evident, that if the base and one of these sides be produced to semicircles, or till they meet, the supplemental A, thus formed, will be isosceles, and can, therefore, be resolved by some of the cases of right-angled triangles, in a similar manner with the former.

MISCELLANEOUS QUESTIONS FOR EXERCISING THE RULES IN THE PRECEDING TABLES.

1. In a right-angled spherical triangle ABC, one of the oblique angles A is 60°, and the other в 45°; required the side B c opposite the former. Ans. Bc 45°.

2. In a right-angled isosceles spherical triangle ABC, the two equal sides AC, CB are each 30°; required the hypothenuse A B. Ans. A B 41° 24′ 35′′. 3. It is required to find the angles of an equilateral spherical triangle A B C, each side of which is 60°. Ans. each 70° 31′ 44′′.

4. In a quadrantal spherical triangle A B C, the hypothenusal angle c is 87°, and one of the other angles A 95° 13'; required the side A c adjacent to that angle. Ans. A c 61° 25′ 53′′.

5. In a right-angled spherical triangle A B C, the hypothenuse A B is 84° 23′ 20′′, and one of the legs A c 96° 36′ 22′′; required the angle A adjacent to that leg. Ans. A 148° 1′ 49′′. 6. In an isosceles spherical triangle ABC, each of the two equal sides AB, AC are 95°, and their included angle 100°; required the base в c. Ans. BC 99° 22′ 24′′.

7. In an oblique-angled spherical triangle ABC, one of its sides A B is 96° 50', the other a c 83° 10′, and their included angle a 120°; required the base BC, and the other two angles в and c.

Ans. B C 120° 28′ 10′′, ▲ B 86° 4′ 13′′,

Lc 93° 55′ 47′′.