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ed to one, viz. that when the thing found by the rules in the first table is ither 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 arch; and this, besides the three ambiguous cases, happens also in other two, viz. the 1st and 11th. The ambiguity is obviated, in these two cases; by this rule, that the sides of a spherical right angled triangle are of the same affection with the opposite angles. Two rules are therefore sufficient to remove the ambiguity in all the cases of the right angled triangle, in which it can possibly be re moved.

It may be useful to express the same solutions as in the annexed table. Let A be at the right angle as in the figure, and let the side opposite to it be a; let b be the side opposite to B, and c the side opposite to C.

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PROBLEM II.

In any oblique angled spherical triangle, of the three sides and three angles, any three being given, it is required to find the other three.

In this Table, the references (c. 4.), (c. 5.), &c. are to the cases in the preceding Table, (16.), (27.), &c. to the propositions in Spherical Trigonometry.

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SOLUTION.

Let fall the perpendicular CD from the unknown angle, not required, on AB.

R: cos. A:: tan. AC: tan. AD, (c. 2.); therefore BD is known, and sin. BD: sin. AD: : tan. A: tan. B, (27.); B and A are of the same or different affection, according as AB is greater or less than BD, (16.)

Let fall the perpendicular CD from one of the unknown angles on the side AB.

R: cos. A tan. AC; tan. AD, (c. 2.); therefore BD is known, and cos. AD: cos. BD: : cos. AC cos. BC, (26.); 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.

3

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4

B.

SOLUTION.

From C the extremity of AC next
the side sought, let fall the per-
pendicular CD on AB.
R: cos AC tan A: cot ACD,

(c. 3.); therefore BCD is known,
and cos BCD: cos ACD :: tan
AC tan BC, (28). BC is less
or greater than 90°, according
as the angles A and BCD are
of the same, or different affec-
tion.

Let fall the perpendicular CD
from one of the given angles on
the opposite side AB.
R: cos AC:: tan A: cot ACD,
(c. 3.): therefore the angle BCD
is given, and sin ACD: sin BCD
:: cos A cos B, (25.); B
and A are of the same or
different affection, according as
CD falls within or without the
triangle, that is, according as
ACB is greater or less than
BCD, (16.)

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and an angle

A

opposite to

one of them,

BC.

The angle

B
opposite to
the other given
side
AC.

The angle
ACB
contained by
the given
sides
AC and BC

The third

side

AB.

Sin BC: sin. AC:: sin A: sin B, (24.). The affection of B is ambiguous, unless it can be det ra mined by this rule, that according as AC + BC is greater or less than 180° A+B is greater or less than 180°, (10).

From ACB the angle sought draw CD perpendicular to AB; then R: cos AU:: tan A: cot ACD, (c. 3.); and tan BC: tán AC:: cos ACD: cos BCD, (28.) 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. 2.); eos AC cos BC:: cos AD: cos BD, (26.)

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ADBD, wherefore AB

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