9-910177

9.437242

+ Cot AC 71° 24'.

9.527065

Angle A is of the same species with BC; and hence A and Care of the same species, therefore (329) AC≤ 90.

+R

Tan BC 140° 53'

Cos C 105° 53'

Cot C 105° 53' +R

+ Sin BC 140° 53′

To find AB.

Comp. BC being M, C and comp. AB are A and a; and hence R sin BC cot C⚫ tan AB; hence

=

Cot C: Rsin BC: tan AB.

+ R
+Sin C 105° 53'

9-799962

Tan AB 114° 17'

10.345814

The side AB and angle C are of the same species (329.)

Cos BC 140° 53'

10.

To find angle A.

Angle A being M, C and comp. BC are O and o; and R cos Asin C⚫ cos BC; and hence R: sin C

•

cos BC: cos A.

10.

Cos A 138° 16'

Angle A is of the same species as BC.

9.454148

10.

9.983094
9.889785

9.872879

EXERCISES.

1. A side and its adjacent angle are respectively 119° 11' and 126° 54'; find the other parts. Ans. The hypotenuse 71° 28', the other side 130° 42', and the other angle 112° 57'.

2. The side AB is 54° 28′ 10′′, and angle A 68° 29′ 48′′; what are the other parts? Ans. AC 75° 19′ 49′′, BC 64° 10′, and C 57° 16′ 1′′.

342. Case VI. When a side about the right angle and its opposite angle are given.

EXAMPLE. Given AB 40° 25′, and angle C 44° 56′; find the other parts.

To find AC.

Comp. of AB being M, AC and C are O and o; and R⚫sin AB sin AC sin C; hence

Sin C: Rsin AB: sin AC.

Sin C 44° 56'

Ꭱ

Sin AB 40° 25'

Sin AC 66° 38'

Or (327) AC is also 180°

Sin A 68° 25'

66° 38'

To find angle A.

Angle C being M, A and comp. AB are O and o; and R cos Csin A cos AB; hence

.

= cos C: sin A.

Cos AB: R
Cos AB 40° 25'
R
Cos C 44° 56'

R

·

Cot C 44° 56'

Tan AB 40° 25'

9-848979

9.811804

9.962825
113° 22′.

Sin BC 58° 36'

Or, BC is also 180°

10.

9.881584

68° 25′ = 111° 35'.

To find BC.

Comp. BC being M, Cand comp. AB are A and R sin BC cot Ctan AB; hence

a;

=

R: cot C

tan AB: sin BC.

10.

9.849990

9.968406

10.
10.001011
9.930220

9.931231

58° 36' 121° 24'.

and

As either of the triangles ABC, A'B'C (fig. to art. 309), fulfil the conditions given in this case, it is hence called the ambiguous case. In practical applications of this subject, it

is generally easily known which of the two triangles is to be taken. If, for example, it were known that the side BC, or the angle A, is less than 90°, the triangle ABC alone would satisfy the given conditions, and the triangle A'B'C would be excluded.

A table containing the solutions of all the cases of this branch of spherical trigonometry is given at p. 68 of vol. II. Geometry; and the reader may if he chooses employ the proportions given in it.

QUADRANTAL SPHERICAL TRIANGLES.

343. If the supplements of the sides and angles of a quadrantal triangle be taken, they will be the angles and sides respectively of a right-angled spherical triangle; the supplement of the quadrantal side being the right angle.

This is evident from the properties of the polar triangle (vol. II. Geom. p. 53); which are the following:

344. If the angular points of a spherical triangle are made the poles of three great circles, another spherical triangle will be formed, such that the sides of each triangle are the supplements of the angles opposite to them in the other triangle.'

F

Thus, if the angular points A, B, C, of the triangle ABC are respectively the poles of the sides EF, DE, DF, opposite to them in the triangle DEF; then EF is the supplement of angle A, DE of B, DF of C, BC of D, AB of E, and AC of F.

N

A

B

G

D

Hence, if ABC were a quadrantal K triangle, AB being the quadrantal side, then DEF would be a rightangled triangle, E being the right angle. Quadrantal triangles can therefore be solved by the rules for right-angled triangles. It will also be sufficient to know two parts of such a triangle besides its quadrantal side; for then two parts of the supplemental right-angled triangle are known, and if its other parts are then calculated, the supplements of its sides and angles will be respectively the angles and sides of the given triangle.

M

H

E

EXERCISE. A side and the angle opposite to the quadrantal side are 136° 8′ and 61° 37′; find the other parts. Ans. The other side 65° 27', and the other two angles 53° 9' and 142° 26'.

OBLIQUE-ANGLED SPHERICAL TRIGONOMETRY.

345. The number of cases in oblique-angled trigonometry, formed in reference to the given parts, is six, as in the former section.

These cases, except when the three sides or three angles are given, can be solved by the method used in the preceding section, as explained afterwards under the next head. The solutions may, however, frequently be more conveniently effected by means of other methods which are here employed for that purpose.

The rules used in the first four cases to determine the species of the part sought are

346. 'The half of a side or angle of a spherical triangle is less than a quadrant.'

For a side or an angle of a spherical triangle is less than two right angles.

Other two rules are given under the fifth and sixth cases, to be employed in their solution.

The rules in article 336 are also applicable to obliqueangled spherical triangles.

347. Half the difference of any two parts of a triangle is less than a quadrant.

For each part is less than 180°.

348. It is to be observed, in forming examples in spherical trigonometry, that the sum of the three sides of a spherical triangle is less than the circumference of a circle; and the sum of any two sides is greater than the third; also the greater angle is opposite to the greater side, and conversely.

349. Case I.-When the three sides are given.

This case can be conveniently solved by any of the three following rules:

I. From half the sum of the three sides subtract the

side opposite to the required angle; then add together the logarithmic sines of the half sum and of this difference, and the logarithmic cosecants of the other two sides; and half the sum will be the logarithmic cosine of half the required angle."

Let the three sides be denoted by a, b, c; the angles respectively opposite to them being A, B, and C; and half the sum of the sides by s, then

sin s⚫ sin (s—a)

sin b⚫ sin c

cos2 A =

; and

Lcos A={L sin s+L sin (s-a)+L cosec b+L cosec c} And for B and C, the formulas are exactly analogous; that for B, for instance, being formed from the above by changing A into B, a into b, and b into a.

II. 'From half the sum of the three sides subtract separately the sides containing the angle; add together the logarithmic sines of the two remainders, and the logarithmic cosecants of these two sides; and half the sum is the logarithmic sine of half the required angle."

If A is the required angle, then

sin2 A

sin (s
sin b sin c

•

L sin A

= { L sin (s — 6) + L sin (s —c) + L cosec b + L cosec c}

tan2 A =

b) sin (sc)

; and

III. From half the sum of the three sides subtract the side opposite to the given angle, and also each of the sides containing it; then add together the logarithmic cosecants of the half sum and of the first difference, and the logarithmic sines of the other two differences; and half the sum will be the logarithmic tangent of half the required angle.' Let A be required, then

sin (s
b) sin (s—c)
; and
sin ssin (sa)
L tan A

Lcosec8+Lcosec (s-a)+L sin (s-6)+Lsin (s-c)} 350. The same remark applies here respecting the circumstances under which the first and second formulas ought