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But, since the tangents are reciprocally as the co-tangents,

Tang AE : tang 3 (AC+BC)::tang 3 (AC-BC): tang DE; and tang 1 (ac+BC) : co-tang 1 (AC~BC)::tang AE : co-tang

DE.

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(T) COROLLARY III. If the triangle be isosceles, or equilateral, the perpendicular will fall on the middle of the base, except the sides be quadrants, and then it may fall in any part of the base.

(U) COROLLARY IV. In a right-angled triangle, the rectangle of the tangents of half the sum and half the difference, of the hypothenuse and one leg, is equal to the square of the tangent of half the other leg. For in this case B and D will coincide, and De will be equal to 1 AB, or equal AE.

PROPOSITION XXVII. (See the Fig. to PROP. XXVI.)

(W) In any spherical triangle,
I. If the perpendicular fall within the triangle.
The co-tangent of half the sum of the angles at the base,
Is to the tangent of half their difference ;

As the tangent of half the vertical angle, Is to the tangent of the excess of the greater of the two vertical angles, made by a perpendicular, above half the aforesaid vertical angle.

II. Or, If the perpendicular fall without the triangle.

The tangent of half the difference between the base angles,
Is to co-tangent of half their sum ;

As the co-tangent of half the vertical angle,

Is to the co-tangent of the ercess of the greater of the two vertical angles, formed by a perpendicular upon the base, above half the aforesaid vertical angle.

DEMONSTRATION. Cos A +COS B : cos A ~COS B::sine ACD + sine BCD : sine AcD~sine BCD. (B. 175.)

Cos A +cos B : cos Arcos B::cot } (A + B) : tang 1 (A~B) (R. 112.) Sine ACD + sine BCD : sine ACD-sine BCD::tang:(ACD + BCD): tang 1 (ACD ~BCD) (P. 111.) Therefore,

Cot ž (A+B) : tang 1 (A~B)::tang 1 (ACD + BCD) : tang (ACD ~BCD.)

But I (ACD + BCD)=} ACB, and } (ACD~BCD)=ECD, hence cot } (A+B) : tang } (A~B)::tang ACB : tang ECD:

But when the perpendicular falls without the triangle, I (ACD - BCD)= ACB, and ECD=} (ACD + BCD.)

Cot į (A~B) : tang (A + B) :: tang 1 (ACD - BCD): tang 1 (ACD + BCD); by inversion, tang 1 (A+B) : cot} (A~B):; tang ECD : tang ACB.

But the tangents and co-tangents are reciprocally proportional.

Tang 1 (A~B) : cot } (A+B):: cot ACB : cot. ECD. Q. E. D.

(X) COROLLARY I. The excess of the greater of the two vertical angles (formed by a perpendicular) above half the vertical angle, is equal to half the difference of those vertical angles, when the perpendicular falls within the triangle; or half their sum, when it falls without. Either of the above rules will bring the same conclusion, whether the perpendicular falls within or without the triangle; only in the former case, the fourth number will be less than half the vertical angle, and in the latter it will be greater. That the rules are the same may be shewn thus,

When the perpendicular falls without the triangle. Tang (A~B) : co-tang Į (A + B)::cot | ACB : cot ECD; by inversion, cot } (A + B) : tang Į (A~B)::cot ECD : cot į

But the tangents are reciprocally as the co-tangents. ::: Cot } (A+B) : tang 1 (A~B): :tang ACB : tang ECD, the same expression as when the perpendicular falls within the triangle.

COROLLARY II. The co-tangent of half the sum of the angles at the base, is to the tangent of half their difference ; as the tangent of half the vertical angle, is to tangent of half the difference between the two vertical angles, formed by a perpendicular, or to tangent of half their sum, according as the perpendicular falls within or without the triangle:

АСв.

CHAP. V.

INVESTIGATION OF GENERAL RULES FOR CALCULATING THE

SIDES AND ANGLES OF OBLIQUE-ANGLED SPHERICAL TRIANGLES WITHOUT MAKING USE OF A PERPENDICULAR.

PROPOSITION

XXVIII.* (Z) If the cosine of any side of a spherical triangle be multiplied by the radius, and the rectangle of the cosines of the other two sides be deducted from the product ; the remainder divided by the rectangle of the sines of these two sides, will be equal to the cosine of the included angle divided by the radius, viz.in any spherical triangle ABC,

(COS AB x rad)-(COS AC X COS BC)

sine Ac x sine BC COS LC

rad

* Legendre's Geometry, 6th Edition, page 386, et seq.

C

COS EDF

COS EOF

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DEMONSTRATION. Let ABC be the triangle proposed, and let o be the centre of the sphere; join ao, BO, and co. Take any point p in

D oc, and in the planes AOC, BOC, draw de and of each at right an

B gles to oc, and join EF.

D Then because the , EDF is the measure of the inclination of the planes aoc, Boc, it is also the measure of the spherical L ACB. (D. 133.)

ED? + DF2 - EF? In the plane triangle EDF,

-(N. 92.) rad

2 DE X DF

EO2 + op? - EF? also in the plane triangle EOF,

rad

2EO X OF From the second of these equations ef?=E02 +0F?280 X OF X COS EOF

which substituted in the first equation gives rad

rad (ED?+DF?-E0%-ora) + (2E0 X OF X COS EOF) COS EDF

2DE X DF But oE – ED' =0D (Euclid 47 of 1.) and oF2 - pro = odo,

(EO X OF X COS EOF)-(od' x rad) consequently cos EDF =

DE X DF

rad Now, L EDF = LC; LEOF=AB;

(O. 44.) =

sine Z DOE rad rad

rad

COS E DOE (O. 44.) sine AC DF

j sine / DOF

sine BC

sine DOE cos / DOF ;

; and if these values be sine AC

OF
sine L DOF

sine BC
substituted in the equation, cos Z EDF=
(rada x cos AB)-(rad x cos AC X cos BC). COS LC
sine AC x sine BC

rad (COS AB x rad)-(cos AC X cos bc)

Q.E.D. sine AC x sine BC (A) Since the preceding conclusion does not depend on any peculiar relation which the 2c has to the other angles, a simiIar equation will be equally true for the angles A and B. Hence

(rad” . cos a)-(rad . cos b. cos c)
Cos LA

sine b. sine c
(rad” . cos b)-(rad . cos cos c)
Cos 6. B

sine a , sine c

OE

DE

OF

OD

DE

COS AC

OD

COS BC

; viz.

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COS a

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(rad” . cos c)--(rad · cos b. cos a). Cos Lo=

sine b. sine a These are the formulæ from which Lagrange and Legendre* begin their investigations, and of the four quantities involved, any three being given the fourth may be found. They are applicable to every species of spherical triangles, whether rightangled, quadrantal, or oblique-angled, but the formulæ for right-angled and quadrantal triangles have already been given. (O. 169, and P. 169.)

(B) If A, B, C represent the three angles of a spherical triangle, the opposite sides may be represented by 180° — A, 180— B, 180°— c; and if a, b, c represent the three sides, their opposite angles may be represented by 180o-- a, 180°-b, 180°-c (U. 137.) Hence Cos (180°-a)=

rad? . cos (180° — A) – rad. cos (180°— e) · cos (180o--c)

sine (180°—B). sine (180° —c) But cos (180°-a) = cos a; cos (180°-A) = cos A, &c. (N. 101.) consequently

(rad? . cos A) +(rad . cos B . cos c)

sine B

sine c In the same manner the cosines of the other sides

may

be determined.

(C) It has been shewn (G. 176.) that the sines of the sides of

any spherical triangle have the same ratio to each other as the sines of their opposite angles ; hence by using the notation of Legendre, we shall have

sine
sine B

sinec
I. Sine A=
sine 0

sinec
sine b. sine A sine b. sine c
Sine B=
sine a

sinec
sinec
sine A

sine c. sine B
Sine c=
sine a

sine 6
sine b. sine A sine c , sine A
II. Sine as

sine B
sine B

sine c. sine B
Sine b =
sine A

sinec
sine a , sine c sine b. sine c
Sine c =
sine A

sine B (D) The general expressions for the cosines, which have been obtained by this proposition, may be arranged thus :

a

sine a •

sine c

sine a

.

* The former in the Journal de L'Ecole Polytechnique, and the latter in his Éléments de Géométrie.

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с

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III. Cos A=

(rad? . cos a),(rad . cos b. cos c)

sine b. sine
(rado . cos b)–(rad . cos a . cose)
Cos B=

sine a

sinec
(rad. cos c) –(rad. cos b. cos a)
Cos c=

sine b. sine a And by reducing these last equations.

(cos A . sine b. sine c)+(rad.cos b. cos c) IV. Cos a=

rado (Cos B. sine a . sine c)+(rad. cos a . cos c) Cos b=

rad?
Cos c.sine b. sine a)+(rad. cos b. cosa)
Cos c=

rad
(rad. cos a)+(rad. cos B. cos c)
V. Cos as

(B. 183.) sine B

sinec
(rad. cos B)+(rad . cos a

cos c)
Cos b=

sine A . sine c
(rad? . cos c) +(rad . cos A . COS B)

sine A . sine B
And by reducing these equations, we shall have

(cos a sine B . sine c)-(rad . cos B cos c) VI. Cos A=

rad (cos b . sine A . sine c) – (rad . cos A . cos c) Cos B=

rade (cosc

sinea . sine B) – (rad . cos A . cos B) Cos c=

rad (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 con

tained 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.

Cos c=

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2

(F) But none of the foregoing formulæ are conveniently adapted to logarithmical calculation.

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