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from this equation C-0 is to be found and then C.

Since C-0 is found from its sine there may be an ambiguity. Again, by Art. 44,

cot a sin cot A sin B=cot a sin c-COS c cos B = cos B

COSC +

9),

cos B

cot a

assume cot

; then

cos B
cot A sin B=cos B (- cosc + sin c cot 0) =

cos B sin (c-6)
sin e

; therefore sin (c - 6) = cot A tan B sin 0;

from this equation c-O is to be found, and then c.

Since c

- A is found from its sine there may be an ambiguity. As before, it may be shewn that these results agree with those obtained by resolving the triangle into two right-angled triangles; for if in the triangle ACB' the arc CD be drawn perpendicular to AB', then B'CD will = 0, and B'D = 0.

86. We now return to the consideration of the ambiguity which may occur in the case of Art. 84, when two sides are given and the angle opposite one of them. The discussion is somewhat tedious from its length, but presents no difficulty.

Before considering the problem generally, we will take the particular case in which a=b; then A must=B. The first and third of Napier's analogies give

cot 1C = tan A cosa, tan} c = tan a cos A ; now cot 1C and tan c must both be positive, so that A and a must be of the same affection. Hence, when a=b, there will be no solution at all, unless A and a are of the same affection, and then there will be only one solution ; except when A and a are both right angles, and then cot }C and tan c are indeterminate, and there is an infinite number of solutions.

We now proceed to the general discussion.

If sin b sin A be greater than sin a, there is no triangle which satisfies the given conditions; if sin b sin A is not greater than

=

= T

sin 6 sin A sin a, the equation sin B =

furnishes two values of B,

sin a which we will denote by B and B', so that '=-B; we will suppose that ß is the one which is not greater than the other.

Now, in order that these values of B may be admissible, it is necessary and sufficient that the values of cot }C and of tan c should both be positive, that is, A - B and a - b must have the same sign by the second and fourth of Napier's analogies. We have therefore to compare the sign of A –ß and the sign of A - B' with that of a - b.

We will suppose that A is less than a right angle, and separate the corresponding discussion into three cases.

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sin a

sin 6 (1) Let a be less than b; the formula sin B= sin A makes B greater than A, and à fortiori B' greater than A. Hence there are two solutions.

(2) Let a be equal to b; then there is one solution, as previously shewn.

(3) Let a be greater than b; we may have then a + b less than a or equal to z or greater than . If a +b is less than then sin a is greater than sin b; thus ß is less than A and therefore admissible, and ß' is greater than A and inadmissible. Hence there is one solution. If a+b is equal to 7, then ß is equal to A, and B' greater than A, and both are inadmissible. Hence there is no solution. If a+b is greater than t, then sin a is less than sin b, and B and B' are both greater than A, and both inadmissible. Hence there is no solution.

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II. Let 6 be equal to

2 (1) Let a be less than b; then B and B' are both greater than A, and both admissible. Hence there are two solutions.

(2) Let a be equal to b; then there is no solution, as previously shewn.

a

(3) Let a be greater than b; then sin a is less than sin b, and B and B' are both greater than A, and inadmissible. Hence there is no solution.

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a

(1) Let a be less than b; we may have then a + b less than a or equal to or greater than t. If a + b is less than then sin a is less than sin b, and ß and ß' are both greater than A and both admissible. Hence there are two solutions. If a +b is equal to , then ß is equal to A and inadmissible, and B' is greater than A and admissible. Hence there is one solution. If a + b is greater than t, then sin a is greater than sin b; B is less than A and inadmissible, and ß' is greater than A and admissible. Hence there is one solution.

(2) Let a be equal to b; then there is no solution, as p.eviously shewn.

(3) Let a be greater than b; then sin a is less than sin b, and ß and ß' are both greater than A and both inadmissible. Hence there is no solution.

We have then the following results when A is less than a right angle.

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It must be remembered, however, that in the cases in which two solutions are indicated, there will be no solution at all if sin a be less than sin b sin A.

In the same manner the cases in which A is equal to a right angle or greater than a right angle may be discussed, and the following results obtained.

When A is equal to a right angle,

<

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T

=

T

!

la<b or a=7

.no solution, b< a>b and a + b < <T

one solution, a>b and a + 6 Tor > TT

.no solution. 6 sa <b or a > 0

.......no solution, 2 Ta= 6

....infinite number of solutions. ra <b and a + b >

.....one solution, b> a <b and a+b = or <

....no solution, a = b or a> 6

.no solution, When A is greater than a right angle, la<b or a=b.......

...no solution, a>b and a+b= or <TT.

one solution, 2 a>b and a + b >

.two solutions. sa <b or a=6.........

.no solution, 2 la> b ......

.two solutions. a <b and a + b >

.one solution, asb and a +6

.no solution, 2 :6

one solution, .two solutions.

T

<

=

T

b =

= T or < T

>

101

a =

a>6

As before in the cases in which two solutions are indicated, there will be no solution at all if sin a be less than sin b sin A.

It will be seen from the above investigations that if a lies between 6 and 7-6, there will be one solution; if a does not lie between 6 and 7-b either there are two solutions or there is no solution; this enunciation is not meant to include the cases in which a=b or = 7

Tb.

87. The results of the preceding Article may be illustrated by a figure.

E

H

A

Let ADAE be a great circle; suppose PA and PA' the projections on the plane of this circle of arcs which are each equal to b and inclined at an angle A to ADA'; let PD and PE be the projections of the least and greatest distances of P from the great circle (see Art. 59). Thus the figure supposes A and b each less than

2

T

If a be less than the arc which is represented by PD there is no triangle; if a be between PD and PA in magnitude, there are two triangles, since B will fall on ADA', and we have two triangles BPA and BPA'; if a be between PA and PH there will be only one triangle, as B will fall on A'H or A H', and the triangle will be either APB with B between A and H, or else A'PB with B be. tween A and H'; but these two triangles are symmetrically equal (Art. 57); if a be greater than PH there will be no triangle. The figure will easily serve for all the cases; thus if A is greater than

2'

we can suppose PAE and PA'E to be equal to A; if 6 is greater than we can take PH and PH' to represent b.

2

T

7T

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