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Or we may treat this case conveniently by resolving the triangle into the sum or difference of two right-angled triangles. From A draw the arc AD perpendicular to CB or CB produced ; then, by Art. 62, tan CD= tan b cos C, and this determines CD, and then DB is known. Again, by Art. 62,

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C=cos AD cos DB = cos DB

; cos CD

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this finds c. It is obvious that CD is what was denoted by 0 in the former part of the Article.

By Art. 62,
tan AD= tan C sin CD, and tan AD= tan ABD sin DB;

thus tan ABD sin DB = tan C sin , where DB=a-6 or 0-Q, according as D is on CB or CB produced, and ABD is either B or the supplement of B; this formula enables us to find B independently of A.

Thus, in the present case, there is no real ambiguity, and the triangle is always possible.

=

tan 1c,

=

sin a

i in

83. Having given two angles and the included side (A, C, B). By Napier's analogies, tan 1 (a+b) =

cos } (A - B)

cos 3 (A + B)
tan} (ab)=
sin 1 (A B)

tano;

sin (4+B) these determine 3 (a+b) and 3 (a-6), and thence a and b,

sin A sin c Then c

may

be found from the formula sin C: this case, since C is found from its sine, it may be uncertain which of two values is to be given to it; the point may be sometimes settled by observing that the greater angle of a triangle is opposite to the greater side. Or we may determine C from equation (3) of Art. 54, which is free from ambiguity,

Or we may determine C without previously determining a and 6 from the formula cos C=-cos A cos B + sin A sin B cos c, This formula may be adapted to logarithms, thus :

cos C = cos B(-cos A +sin A tan B cos c); assume cot p=tan B

COS C;
then

cos B sin (4-6).
cos C=cos B(- cos A + cot & sin A) =

; this is adapted to logarithms,

Or we may treat this case conveniently by resolving the triangle into the sum or difference of two right-angled triangles. From A draw the arc AD perpendicular to CB (see the righthand figure of Art. 82); then, by Art. 62, cos c=cot B cot DAB, and this determines DAB, and then CAD is known, Again, by Art. 62,

cos AD sin CAD= cos C, and cos AD sin BAD=cos B;

+

sin

cos C

cos B

therefore

this finds C. sin CAD

sin BAD; It is obvious that DAB is what was denoted by $ in the former part of the Article,

T. S. T.

E

By Art. 62, tan AD = tan AC cos CAD, and tan AD = tan AB cos BAD; thus

tan b.cos CAD = tan c cos , where CAD= A -o; this formula enables us to find 6 independently of a.

Similarly we may proceed when the perpendicular AD falls on CB produced ; (see the left-hand figure of Art. 82).

Thus, in the present case, there is no real ambiguity; moreover the triangle is always possible.

84. Having given two sides and the angle opposite one of them (a, b, A). The angle B may be found from the formula

sin b sin B=

sin A; sin a

and then C and c may be found from Napier's analogies,

tan 1C =
cos 1 (a-6)

cot! (A + B),
cos (a + b)

cos } (A + B)
tan c=
1

tan 1 (a + b).

cos (A - B) In this case, since B is found from its sine, there will sometimes be two solutions; and sometimes there will be no solution at all, namely, when the value found for sin B is greater than unity. We will presently return to this point. (See Art. 86.)

We may also determine C and c independently of B by formulæ adapted to logarithms. For, by Art. 44,

cot A cot a sin b = cos b cos C + sin C cot A = cos b (cos C +

cot A assume tan $=

cos 6 cos (C ");

( cot a sin b = cos 6 (cos C + tan o sin C) =

cos o therefore cos (C - $) = cos o cot a tan b;

sin C);

cos 6

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cos 6

; thus

ز

from this equation C-is to be found, and then C. The ambiguity still exists; for if the last equation leads to C -0= a, it will be satisfied also by 0-C=a; so that we have two admissible values for C, if $ + a is less than 7, and $-a is positive.

And cos a = cos 6 cos c+sin b sin c cos A = cos 6 (cos c + sin c tan b cos A); assume tan 0= tan b cos A ; thus

cos b cos (c-0). cos a = cos 6 (cos c+sin c tan 6) =

;

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from this equation c-O is to be found, and then c; and there may be an ambiguity as before.

Or we may treat this case conveniently by resolving the triangle into the sum or difference of two right-angled triangles.

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=

Let CA = b, and let CAE = the given angle A ; from C draw CD perpendicular to AE, and let B and CB' = a; thus the figure shews that there may be two triangles which have the given elements, Then, by Art. 62, cos b=cot A cot ACD; this finds ACD, Again, by Art. 62,

tan CD= tan AC cos ACD,

and tan CD= tan CB cos BCD, or tan CB' cos B'CD, therefore tan AC cos ACD= tan CB cos BCD, or tan CB' cos B'CD; this finds BCD or B'CD.

It is obvious that ACD is what was denoted by $ in the former part of the Article,

COS

Also, by Art. 62, tan AD= tan AC cos A ; this finds AD. Then

l AC:

= cos CD cos AD,

CB = cos CD cos BD, or cos CB' = cos CD cos B'D;

COS

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It is obvious that AD is what was denoted by 0 in the former part of the Article,

85. Having given two angles and the side opposite one of them (A, B, a).

This case is analogous to that immediately preceding, and gives rise to the same ambiguities. The side b may be found from the formula

sin B sin a sin b =

;

sin A and then C and c may be found from Napier's analogies,

cos } (a - b)
tan} C =

cot } (A + B),
cos } (a + b)
tan
cos} (A + B)

tan} (a+b).

cos (A - B) We may also determine C and c independently of l by formula adapted to logarithms. For

C

cos

A=- =-cos B cos C + sin B sin C cos de

= cos B(- cos C'+ tan B sin C cos a), assume cot $ = tan B cos a ; thus = cos B (- cos C + sin C coto)

cos B sin (C – ");

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sin •

therefore

sin (C-6)=

cos A sin

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cos B

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