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cota.ro cot a . tang b Secco

sine (6-C) and tang c=r Also, Tang La=

sine (c+b)

cot 6

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tang b.r
tang a
sine(b-a)

sine (6+a)

cosecc

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CASE V. Given the hypothenuse and a leg, to find the angle opposite to that given leg.

sine cir
sine c . cosec b

cosec bir SOLUTION. Sine că

sine b
cosecc.

sine bir sine b.cosecc
Cosec c=
cosec 6

sinec
sine a . cosec b

cosec 6.9 And, Sine A=

sine b

sine 6.4 sine b. cosec a Cosec A=

cosec 6 Also, tang (45° + } c) = £ v tang 1 (6+c). cot } (6-c). And, tang (45° + A) = I tang 1 (b+a). cot }(b-a).

sine a.1

cosec a

coseca.7

sine a

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CASE VI. Given the hypothenuse and one side or leg, to find the other leg.

cos b.gro cos b. secc SOLUTION. Cos a =

secc

.

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COSC

sec 6

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seci b.rs

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Sec a =

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cos 6.7 And, Cosc =

COS a

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sec bor

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Secc

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Also, tang į a = tang 1 (b+c). tang 1 (6c).
And, tang 1 c = v tang 1 (6+a). tang 1 (b-a).

CASE VII. Given a side of a right-angled spherical triangle, and the angle adjacent, or joining to it, to find the side opposite to the given angle. sinec

tang a SOLUTION. Tang a

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CASE VIII. Given a side of a right-angled spherical triangle, and its adjacent angle, to find the angle opposite to the

given side.

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cos c. sin a

COS C re sin А SOLUTION. Cos c =

r

cosec A secc cosec A. cosec A. secc secc.na Sec c = COS C

n

sin A cos a , sine c COS at

sine cir And, Cos A =

cosecc sec a cosecc.r cosec c. sec a

seca. r Sec A = cos a

1

sinec

COS AT

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sec A

r

sec A

r

CASE IX. Given a side and its adjacent angle, to find the hypothenuse.

cot c . COS A cot con SOLUTION. Cot b =

tang 4 tang c

sec A . tang c Tang b =

cot c
COS C. 1 cot a . cos C cot a
And, Cot b =

tang a

tang a . sec C. sec c . tang a Tang b =

cot a

COS A

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secc

COS C

r

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Case X. Given a side of a right-angled spherical triangle, and its opposite angle, to find the side adjacent to that angle. SOLUTION.

Sine c =
tang a . cot a

tang aircot Air
tang A

cot a
tang Air
Cosec c =

tang A . cota

cot a r tang a

cot A

cot c.19
tang c • cot c_tang c :*
And, Sine a =

tang C

cot c
tang cur tang c. cot c cot c.r
Cosec a =
tangc

cot c

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Also, Tang (45°+$e)=+rV sine (A+a)

sine (A-a) And, Tang (45o+ļa)= + V sine (C+c)

sine (c-e)

COS A

sec a

seca.ro

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sec A

sec A

COS a

COS a

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COS A

CASE XI. Given a side of a right-angled spherical triangle, and its opposite angle, to find the adjacent angle.

COS A.19 SOLUTION. Sine c=

COS a

sec A.1 Cosec c=

sec a

COS C.ro
And, Sine as

COS C

sec c.r Cosec A=

secc Also, Tang (45° ++ c) = + cot } (n+a). cot } (A-a). And, Tang (45° +1A) = † / cot (c+c). cot $ (C-c).

COS C. secc

secc.19

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sec C

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CASE XII. Given a side of a right-angled spherical triangle, and its opposite angle, to find the hypothenuse.

sine.a. sine a . cosec A SOLUTION. Sine b=

sine A sine A .r

sine A . cosec a Cosec b

sine a sine c.ro

sine . cosecc And, Sine b=

sine c

sine c.r sine C. cosecc Cosec b=

sine c Also, tang (45° + 1b)=+ Vtang 1(c+c). cotě (c-c). And, tang (45° +16=tang 1 (A +a). cot 1 (A-a).

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CASE XIII. Given the two sides, or legs, of a right-angled spherical triangle, to find an angle.

sinec

sine c.

cot a SOLUTION. Cot A

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CH

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GEN

CASE XIV. Given the two sides, or legs, of a right-angled spherical triangle, to find the hypothenuse.

SOLUTION. Cos b=

COS a . COS C

COS a.ro

COS C.7

r

secc

sec a

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T

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CASE XV. Given two angles of a right-angled spherical iangle, to find a side, or leg.

COS A . cosecc Solution. Cos a=

sine c sine c

sine c. sec A sec A.19 Sec a

COS A.1

cosecc.1

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sec A

.

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CASE XVI. Given two angles, of a right-angled spherical triangle, to find the hypothenuse.

cot c. cotA

cot Å.90 cot c.ge SOLUTION. Cos b=

tang C tang A
tang cor

tang c. tang A_tang Air
Sec b=
cot A

cot c

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cot A, cotc cot c.ro cot Air
And, Cos b=

1°
tang A

tang C
Sec b=

tang A.r tang A . tang c_tang c.r
cot c

cot A
cos (a + c)
cos (A~c)

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Also, tang 6=V

a

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GENERAL OBSERVATIONS ON THE SPECIES AND AMBIGUITY OF

THE CASES. (A) The species of the sides and angles may be determined from the equations produced by Baron Napier's Rules, or from the preceding formulæ, by attending to the signs of the quantities which compose the equations or formula.

The sides which contain the right angle are each of the same species as their opposite angles, viz. a is of the same species with A, and b is of the same species with B. (R. 145.)

It may be proper to observe that where a quantity is to be determined by the sines only, and a side or angle opposite to the quantity sought does not enter into the equation, the case will be ambiguous, thus in the with case, where sine b = sine

the hypothenuse b is ambiguous. sine A

sine A . sine b Again, in the vid case, where sine a=

the sine of a is evidently determinate, because it is of the same species with a which is a given quantity.

(B) When an unknown quantity is to be determined by its cosine, tangent, or cotangent, the sign of this value will always determine its species ; for, if its proper sign be+, the arc will be less than 90°; if the proper sign be-, the arc will be greater than 90°. (K. 100.)

(C) Again, in Case vith, where rad x cos b=cos C x cos a, it is obvious that the three sides are each less than 90°, or that two of them are greater than 90°, and the third less; as no other combination can render the sign of cos c x cos a like that of cos b as the equation requires. *

QUADRANTAL TRIANGLES. (D) Any spherical triangle of which A, B, C, are the angles, and a, b, c, the opposite sides, may be changed into a spherical triangle of which the angles are supplements of the sides a, b, c,

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* Legendre's Geometry, 6th Edition, page 381.

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