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And by reducing these last equations. (cos A. sine b. sine c)-1-(rad. cos b. cos c) IV. Cos az- e

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And by reducing these equations, we shall have (cos a . sine B. sine C)–(rad . cos B. cos c) VI. Cos A=

Cos b

Cosci

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(E) The six preceding articles afford solutions to all the different cases of oblique-angled spherical triangles. The Ist. Finds the angles, when two sides and an angle opposite to one of them are given. The IId. Finds a side, when two angles and a side opposite to one of them are given. The IIId. Finds the angles, from the three sides being given. The IVth. Finds the third side, when two sides and their contained angle are given. The Vth. Finds the sides, from the three angles being given. The VIth. Finds the third angle, when two angles and the side adjacent to both of them are given. (F) But none of the foregoing formulae are conveniently adapted to logarithmical calculation. .

Let the value of the cosine of c(D. 183.) be substituted in the formula rad”—rad. cos c=2 sine” + c (2d equation I. 117.) we

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- sine a. sine b ;but(sinea. sineb)

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hence sine # c-rad Vincia:Fo +c)—b.sine{(a +b+c)-a, sine a . sine b

and it is evident that the same formula will be obtained, with

only a change of the letters, for the angles A and B.

• When # Z. c is near 90° this will not be a convenient rule for producing an accurate result, because the difference of the logarithmical sines for 1" is then very small (see the note page 54); if Z c be less than 45°, it will be proper to use this rule.

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* Eléments de Géométrie, 5th edition, page 891, et seq.

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therefore

* Though the quantity under the radical sign appears under a negative form, it is always positive; for, since the three angles of every spherical triangle are together greater than two right angles (T. 157.)3 (A+B+c) must be greater than 30°, and the cosine of an arc greater than 90°, is negative, (K. 100,) therefore the expression—cos 4 (A+B+c) becomes positive. The other part under the radical sign is always positive; for, if the supplements of the three angles A, B, c be taken, they will give the three sides of a new triangle = 180°–A, 180°–B, 180°–c, (U. 137.) any two of which taken together are greater than the third (Y. 138); hence 180°–A is less than 180°-54-180°–c, that is, – A + b + c is less than 180°, consequently 4 (B+ c-A) or its equal # (A + i + c)-A is less than 90°, and the cosine of an arc less than 90° is always affirmative. (K. 100.) Legendre, page 392.

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(L) Also, because rad. a. =tang # a (N. 104.) and

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andcolo-nav/o #(or 9- e coso (A+B+C)-B. —cos # (A+B+C). cos # (A+B+C)-A And by the same process a similar formula will be obtained for the other sides. Also for the reasons already given (Note I. 187.) the quantities which appear with a negative sign are positive. Any one of the three preceding formulae (I. 186. K. 187. L. 188.) will determine a side when the three angles are given. (M) From the third set of equations (D. 183.) we have Cos A. sine b. sine c-rado. cosa-rad . cos b. cos c, and cos c. sine b. sine a-rad”. cos c-rad . cos b. cos a. By exterminating cos c and reducing the equations, we get rad . cos A. sine c-(rad. cos a . sine b)–(cos c. sine a , cos b.) This last equation, by a simple permutation, gives rad. cos B. sine c-(rad . cos b. sine a)–(cos c. sine b. cosa).” By adding the last two equations together, and reducing them we obtain sine c. (cos A+cos B)=(rad—cos c). sine (a+b).

... sine c sine a sine b ... But since: e a S (C. 183.) we have

sine cTsine ATsine B sine c. (sine A+sine B)=sine c. (sine a +sine b). And sine c. (sine A-sine B)=sine c. (sine a-sine b). Dividing successively these two equations by the preceding equation, we obtain

= cot # a (O. 104.) we shall obtain by division

* Legendre's Geometry, page 394.

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