a

саа

Case 3. Data : Two sides and the angle opposite one of

them.
Given a, b, A; Sought c, B, C.

6
Sin B =

sin A; C = 180° – A – B;

;
sin C sin C

siu A sin B Note.-_When the side 6 adjacent to the given angle A is greater than a the side opposite to it, there are two triangles which furnish the same data a, b, A; and, consequently, the solution is ambiguous. Case 4. Data : Two angles and the side between them. Given A, B, C; Sought a, b, C.

sin A

sin B C = 180° - A - B; a = 0 6

sin

sin C

III, RIGHT-ANGLED SPHERICAL TRIANGLES,

А

B

с

a

In the triangle A B C, let C be the right angle and c the hypothenuse. Case 1. Data : The two sides. Given a, b; Sought c, A, B.

tan 6 cos c = cos a cos b; tan A = ; tan B =

sin 6

sin a

tan a

Case 2. Data : The hypothenuse and a side.
Given c, b; Sought a, A, B.

tan 6

sin 6 ; cos A =

sin B =

3 tan o

sin

COS c
COS a =

Casę 3. Data : A side and the angle opposite.

Given a, A; Sought b, c, B.
sin b = ; sin c=

; sin B =

tan a

sin a

cos A

tan A

sin A

COS a

Note. — This case is ambiguous, each of the things sought having two values, viz. either the angle found or its supplement. Thus, sin b = sin (180° — b).

Case 4 Data: A side and the adjacent angle.

Given a, B; Sought b, c, A. tan b=sin a tan B; cotc=cot a cos B; cos A = cos a sin B.

Case 5. Data : The hypothenuse and an angle.

Given c, A; Sought a, b, B. sin a = sin c sin A ; tan b = tan c cos A ; cot B = cos c tan A,

Case 6. Data : The two oblique angles.

Given A, B; Sought a, b, c,

cos A
cos a =

cos B
; cos 6 =
sin B

sin A

; cos c = cot A cot B.

Case 1. Data : The three sides.

Given a, b, c; Sought A, B, C.
Let s =

1 (a + b + c);
Assume M =
(sin (s – a) sin (s – 6) sin

(s

sin (s

then,

M tan ; A =

M tan + B =

sin (s – 6) :

M tan įC=

sin (s

c)
Case 2. Data : Two sides and the included angle.

Given a, b, C; Sought c, A, B.
Find an angle o such that

cot Q = tan a cos C;
then,

cos a sin (6 + 0)

COS C =

sin P

sin a sin c

sin A = sin C

;

sin c

=

sin 6
sin B = sin C
Otherwise,
tan 4 (A + B)

cot } C cos } (a - b);

cos } (a + b) tan (A - B)

cot } C sin } (a - b)

;

sin ; (a + b)
A=} (A + B) + 1 (A - B);
B= 1 (A + B)- 1 (A - B);
sin C

sin C
sin c = sin a

sin A

= sin 6

sin Bo

Case 3. Data : Two sides and the angle opposite one of

them.
Given a, b, A; Sought c, B, C.
Find two angles Q and of such that

= tan b cos A ; tan = cos b tan A;
then,
sin (c + 0)
cos a sin Q, sin B = sin A

sin b
sin (C++) = cot a tan b sin 4.

cot

Case 4. Data : Two angles and the side between them.

Given A, B, c; Sought a, b, C.
Find two angles Q and 4 such that
tan PE

= cos c tan A; tan = cos c tan B;
then,
tan c sino

tan c sin 4
tan a =

; tan b =
sin (B + 0)

sin (A + )
cos A cos (B + ®) cos B cos (A + Y)
cos C =

cos y
Case 5 Data : Two angles and the side opposite one of

them.
Given A, B, a; Sought b, c, C.
Find two angles 9 and 4 such that

tan Q = tan a cos B ; cott = cos a tan B;
then,

sin a sin B

;

sin A sin (c - 0)

= cot A tan B sin Ø;

cos A sin 4
sin (C-4)
Note. This case is ambiguous.
Case 6. Data : The three angles.

Given A, B, C; Sought a, b, c.

Let S= (A + B + C);
Assume N =

(S – A) cus (S – B) cos C)
then,

tan da= N cos (S – A);
tan + b N cos (S – B);
tan c= N cos (S – C).

sin 6 =

cos B

cos S

Let F = the area or superficial content of a spherical
triangle, r = radius of the sphere, and g = the semicircum-
ference to radius 1. Then,

F
A + B + C - 180°

22

CONSTANTS.

Logarithm

7

Ratio of the circumference of a circle to ? its diameter .....

3:1415927 0°497150 go? 9.8696044 0.994300 god 31.0062767 10491450 N« 1'7724539 0·248575

V r 1'4645919 0:165717

Hyperbolic log. a I'1447299 0'058703 Area of circle, diameter = 1.......... 0°7853982 9.895090

4

Content of sphere, diameter = 1.......:

0°5235988 19*718999

Diameter of circle, area = 1 ..........

Diameter of sphere, content = 1......

VE
V12407010 0093667

} 0-00000485 4:685575
} 0-00029089 6-463726

Length of arc 1' (= sin i') in parts of the

radius.... Length of arc 1' (= sin 1') in parts of the

radius ....... Length of arc 1° in parts of the radius.... Sin ro in parts of the radius Radius reduced to seconds .. Radius reduced to minutes.. Radius reduced to degrees... 360 degrees expressed in seconds.

oʻ01745329 8.241877 0.01745241 8.241855 206264081 5'314425 3437*7468 3.536274 57'295780 1°758123 1296000 6°112605

Number whose hyperbolic log. is unity..
Modulus of the common logarithms ...
Complement of same

2.7182818 0'434294 0°4342945 9637784 2°3025851 0'362216