tan E=

tanza.tan b.sin C
1 + tanza.tanb.cos C

sinja.sin b.sin C
4 cos;($+4).cos; (0-0)
if cos (=sinļa.sin b.(cos}C)?,

and cos o= cos(a + b). (W. Ch. viii. prop. 13; Leg. Geom. Note x; L. 177, 263—6.) (24.) Let R, R, be the circular radii of the inscribed, and circumscribing circles,

N
2 cos į A.cosįB.cos C
2 sinşa.sinžb.sin şc

(L. 270, 1.) N

n

tan R

sin s

COSS

tan R,

n

(25.) Solution of right-angled spherical triangles.
Let A, B, be the oblique angles, C, the right angle ;

a, b, the sides, C, the hypothenuse.
Naper's Rules. The circular parts are

90° - A, 90°-B, a, b, 90° - C; any one of which being called the middle part, M, the two adjacent to m on each side of it, A1, A2, and the two remaining or opposite parts, 01, 02 ;

sin m=tan 17. tan

دوم

= COS 01.COS 02.

[1] cos c=cot A.cot B;
[2] cos c=cos a.cos b;
[3] sin a=tan b.cot B;
[4] sin a=sin c.sin A;
[5] cos A= tan b.cot c;
[6] cos A=cos a.sin B.

These are all the forms essentially different ; four more analagous to [3], [4], [5], [6], may be obtained by changing A, a, into B, b, and vice versa.

These rules may be applied to a quadrantal triangle, in which c=90°, if

A, B,

- (90°-C), 90°-a, 90° - b, be taken as the circular parts.

Any angle, and the side opposite are either both > 90°, or both < 90°.

An oblique angle cannot be less, if acute, nor greater, if obtuse, than the opposite side.

The sides are both > 90°, or both < 90°, if the hypothenuse <90°; and one side > 90°, and the other < 90°, if the hypothenuse > 90°.

A side is > or < hypothenuse, according as it is > or < 90°.
A + B and a + b are both >, or both 180°.
A + B is always > 90°, and AB always < 90°.

(W. Chap. X; L. 202—9.) [1] Given A, c: sin a=sin c.sin A ; log Sin a=log Sin c + log Sin A - log r.

tan b=tan c.cos A ; log Tanb=log Tan c + log Cos A – logr.

cot B=cos c.tan A; log Cot B=log Cosc + log Tan A – log r.

<,

The results obtained in [2], [4], are doubtful; the ambiguity may in some cases be removed by attending to the conditions stated in page 118.

When a small angle is to be determined from its cosine, or an angle nearly = 90° from its sine, a small error in the sine or cosine gives a large one in the angle; in such cases some of the following formulæ will give more accurate results.

tan ( A + B - 459)
tanţa=

tan ( A - B - 459)
tan ža=tan (c +b). tan (c-b)|*.

1 2 3
sin (c-6)
tanA=

sin (c+b)
tan (45° — ļa)=tan (45° — ~)*; if tan x=sin c.sin A.

13

tan; (c + a) tan (45° +į A)=

tan (c-a)

tan (A + a) tan (45° +èc)=

tan}(A — a)

sin (A + a) tan (45° +5)=

sin (A-a

)

cot (A + a)
tan (45°+B) :

tan; (A - a)
Formulæ adapted to a table of natural sines.
Sina = Cos (c- A) - Cos (c + A).
Cosc = Cos (a + b) + Cos (a — ).
Cos A= Sin (a + B)-] Sin (a – B).

(W. Ch. x; L. S. v; C. Ch. xvii.)

– 6

;

(26.) Solution of oblique-angled spherical triangles. First Case: given a, b, c.

2 [1] sin A= . sin 8. sin (8—a). sin (s— b), sin(8—c))";

sin b.sin c log Sin A=log 2 + log r - log Sin b- log Sin c +}{log Sin 8 + log Sin (8-a) + log Sin (s—b) + log Sin (s — c)}.

sin (8—b). sin (8 —c) [2] (sin A)' =

sin b.sin c log Sin A=}{log Sin (s-- ) + log Sin (8--c)

log Sin b-log Sinc} + log r.

sin s.sin (s - a) [3] (cos A)2 =

sin b.sin c log Cosz A= {log Sin s + log Sin (s-a)

– log Sin b-log Sinc} + log r.

sin (s—b). sin (s—c). (tanja)=

sin s.sin (s - a) log Tanza=}{log Sin (8—6) + log Sin (s—c)

- log Sin s - log Sin (s-a)} + log r. [5] Assume cos = cos b.cosc, log Cos 0 - log Cos b + log Cosc – log r; then

2 sin}(@+a).sin:((-a)

sin b. sin c log Cos A=log 2 + log Sin}(@+a) + log Sinį(0 - a)

– log Sinb – log Sin c + log r. A-90°, and e- a have the same sign.

cosb.cos C [6] Assume tan p=

ܪ

COS' A :

;

sin a

log Tan p=log Cosb + log Cosc – log Sin a; then

cos (a +0) COS A =

cos p.sin b.sinc log Cos A=log Cos (a +0) + 3 log ro – log Cosp-log Sin b— log Sin c.

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