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 bo 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, viz. the 1st and 11th. The ambiguity is obviated, in these two cases, by this rule, that the sides of a spherical right angled tri angle 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 removed.

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

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.

and the in

Let fall the perpendicular CD from
the unknown angle, not requir-
ed, on AB.

One of the R: cos A:: tan AC: tan AD,
(c. 2.); therefore BD is known,

B.

other angles 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.

2 cluded angle The third 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.

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, (28.) ACD|
± BCD = ACB, and ACB is
ambiguous, because of the am-
biguous 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.); cos AC: cos BC:: cos
AD: cos BD, (26.)
AB=ADBD, wherefore AB
is ambiguous.