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adjacent to it, adjacent parts; and the two remaining parts, opposite parts; then, the rectangle under the radius and the cosine of the middle part, is equal to that under the cotangents of the adjacent parts, or the sines of the opposite parts. Or if the middle part be called M; the two adjacent parts, A and a; and the opposite parts O and o,

Rcos Mcot A · cot a,

or R cos Msin Osin o.

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Either of these rules may be converted into a proportion by Pl. Ge. VI. 16.

The cases marked ambiguous are those in which the thing sought has two values, and may either be equal to a certain angle, or to the supplement of that angle. Of these there are three, in all of which the things given are a side, and the angle opposite to it; and accordingly, it is easy to show that two right-angled spherical triangles may always be found, that have a side and the angle opposite to it the same in both, but of which the remaining sides, and the remaining angle of the one, are the supplements of the remaining sides and the remaining angle of the other, each of each.

Though the affection of the arc or angle found may in all the other cases be determined by the rules in the second of the preceding tables, it may be useful to remark, that all these rules, except two, may be reduced to one, namely, that when the thing found by the rules in the first table is either a tangent or a cosine; and when, of the tangents or cosines employed in the computation of it, one only belongs to an obtuse angle, the angle required is also obtuse.

Thus, in the 15th case, when cos AB is found, if C be an obtuse angle, because of cos C, AB must be obtuse; and in case 16, if either B or C be obtuse, BC is greater than 90°, but if B and C are either both acute, or both obtuse, BC is less than 90°.

It is evident that this rule does not apply when that which is found is the sine of an arc; and this, besides the three ambiguous cases, happens also in other two, namely, the 1st and 11th.

Solution of the Cases of Oblique-Angled Spherical
Triangles.

PROBLEM.

In any oblique-angled spherical triangle, of the three sides and three angles, any three being given, the other three may be found.

In this table, the references (c. 4), (c. 5), &c. are to the cases in the preceding tables.

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SOLUTION.

Let fall the perpendicular CD
from the unknown angle
not required, on AB.
R: cos Atan AC: tan AD
(c. 2); therefore BD is
known, and sin BD:sin
AD: : tan A: tan B (10);
B and A are of the same
or different affection, ac-
cording as AB is greater or
less than AD (Sp. Ge. 16).

Let fall the perpendicular CD
from one of the unknown
angles on the side AB.
R: cos Atan AC: tan AD
(c. 2); therefore AD is
known, and cos AD: cos
BD::cos AC: cos BC (9);
according as the segments
AD and DB are of the
same or different affection,
AC and CB will be of the
same or different affection.

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SOLUTION.

Sin BC: sin AC:: sin A: sin

B (7). The affection of B is ambiguous, unless it can be determined by this rule, that according as AC+BC is greater or less than 180°, A+B is greater or less than 180° (Sp. Ge. 10).

From ACB, the angle sought, draw CD perpendicular to AB; then

R:

: cos AC :: tan A: cot ACD (c. 3); and tan BC: tan AC cos ACD: cos BCD (11). ACD + BCD = ACB, and ACB is ambiguous, because of the ambiguous sign + or

Let fall the perpendicular CD from the angle C contained by the given sides upon the side AB.

R: cos A:: tan AC: tan AD (c. 5); cos AC: cos BC:: cos AD: cos BD (9). AB = AD ± BD; wherefore AB is ambiguous.

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SOLUTION.

Sin B: sin A:: sin AC: sin BC (7); the affection of BC is uncertain, except when it can be determined by this rule, that according as A+B is greater or less than 180°, AC + BC is also greater or less than 180° (Sp. Ge. 10).

From the unknown angle C draw CD perpendicular to AB; then

R: cos A::tan AC: tan AD (c. 3); tan B: tan A :: sin AD: sin BD. BD is ambiguous, and therefore AB = AD + BD may have four values, some of which will be excluded by this condition, that AB must be less than 180°.

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