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RUL E II.
the middle Part, is equal to the Reclangle un.
Each of the Rules have three Cases: For the middle Part may be the Complement of the Angle B, or C; or the Complement of the Hypothenuse BC; or one of the Legs, A B, A C.
Cafe 1. Let the Complement of the Angle C be the middle Part; then shall AC, and the Complement of the Hypothenuse B C, be adjacent Extremes. By Prop. 28. the Cofine of the vertical Angle C is to. Radius as the Tangent of C A is to the Tangent of the Hypothenuse BC. Then (by Alternation) we shall have Cor. C:T,CA::R:T, BC. But R:T, BC:: Cot. BC:R (as has been before shewn). Wherefore Cof. C:T, AC:: Cot. BC:R; whence R Cof. C=T, A Cx Cot. B C.
And the Complement of the Angle B, and AB, are opposite Extremes to the same middle Part, the Complement of the Angle C; and (by Prop. 25.) as the Cofine of the Angle C, to the Sine of the Angle CDF, fo is the Cofine of DF to Radius. But the Sine of CDF=A E=Cor. B A, and Cor. DFS, EF=S, Angle B. Whence it will be, as Cor. C: Cof. BA::S, B: R. And R+Cor. C=Cof. BA XS, B; that is, Radius drawn into the Sine of the middle Part, is equal to the Rectangle under the Cofines of the opposite Extremes.
Cafe 2. Let the Complement of the Hypothenuse BC be the middle Part; then the Complements of the Angles B and C will be adjacent Extremes. In the Triangle DCF (by Prop. 27.) it is, as S, CF:R:: T, DF: Tic. Whence (by Alternation) S, CF: T, DF::(R:T, C::) Cot. C:R. But S, CF =Cof. B C and T, DF=Cot. B. Wherefore RX Col. B C-Cot. Cx Cot. B ; that is, Radius drawn into the Sine of the middle Part is equal to the Product of the Tangents of the adjacent extreme Parts. X 3
And B A, AC, are the opposite Extremes to the said middle Part, viz. the Complement of BC; and (by Prop. 26.) Cor. B A : Cof. BC::R: Cor. A C. Wherefore we shall have R x Cor, BC=Cof. BA= Cor. A C.
Case 3. Lastly, let A B be the middle Part; and then the Complement of the Angle B, and A C, will be adjacent Extremes, and (by Prop. 27.)S, AB: R::T,CA:T, B. Whence S, AB:T, CA:: (R:T, B::) Cot. B: R. And so R+S, AB=T, CA+Cot. B.
Moreover, the Complements of BC, and the Angle C, are oppofite Extremes to the same middle Part AB; and in the Triangle GHD (by Prop. 25.) we have Cor. D:S, DGH:: Col. GH:R. But Cor. D=Cof. A E=S, A B, and S, G=S, IF-S, BC. Also, Cor. GHES, HI=S, C. Wherefore it will be, as S, AB:S, BC::S, C:R. And hence R XS, A B=S, B C#S, C.
And so in every Case, the Rectangle under the Radius, and the Sine of the middle Part, shall be equal to the Rectangle under the Cofines of the opposite Extremes, and to the Rectangle under the Tangents of adjacent Extremes: And, consequently, if the aforesaid Equations be resolved into Analogies (by 16 EI, 6.) the unknown Parts may be found by the Rule of Proportion. And if that part fought be the middle one, then shall the first Term of the Analogy be Radius, and the second and third, the Tangents or Cofines of the extreme Parts. If one of the Extremes be sought, the Analogy must begin with the other; and the Radius, and the Sine of the middle Part, must be put in the middle Places, that so the Part sought may be in the fourth Place.
N Oblique-angled Spherical Triangles (Fig. ta
Prop. 31.) BCD, if a perpend:cular Arc AC be let fall from the Angle C to the Base, continued, if need be, so as to make two Right-angled Spherical Triangles B AC, DAC; then, by those Right-angled Triangles, may most of the Çales of Oblique-angled oncs be solved,
BD, are proportional 10 the Sines of the verti
cal Angles B CA, DCA.
OR Cf. A g'e B : S, BCA:: (Cor. CA:R
procal Proportion of the Tents of the Angles
T, of the Angle B. And by the same inversly, R:S, DA::T, of the Angle D:T, A C. Then will it be by the Equality of perturbate Ratio, according to Prop. 23. El. 5.) S, BA:S, DA::T, Angle D:T, Angie B.
PROPOSITION XXXIV. The Tangents of the Sides BC, DC, are in a reciprocal Proportion of the Gofines of the vertical
Angles BCA, DCA.
T, BC:R::T, CA: Cor. BCA, and by the fame, R: Cor. DCA::T, DC:T, CA. Wherefore, by Equality of perturbate Proportion, T, BC: Cof. DCA::T, D'C: Cof. BCA.
PROPOSITION XXXV. The Sines of the Sides B C, DC, are proportional
to the Sines of the opposite Angles D and B. Ecause (by the 29th of this) S, BC:R::S, CA:
S, of the Angle B; and, by the same, inverting, -R:S, DC::S, Angle D: S, of C A. Whence, by Equality of perturbate Ratio, S, BC:S, DC::S, D:S, B.
C FRA E, or FM XA E, contained under ibe
B; and let B P, BN, be Quadrants; and then
also, describe from the same Pole B a lesser Circle CFM thro'C; the Planes of these Circles shall be perpendicular to the Plane BON (by the ad of this.) And PG, CH. being perpendicular in the same Plane, fall on the common Sections ON, FM; suppose in G, H. Again, draw H I perpendicular to A 0; and then the Plane drawn thro' CH, HI, shall be perpendicular to the Planè A OB. Whence A I, which is perpendicular to HI, will be perpendicular to the Right Line CI (by Def. 4. El. 11.); and fo A I is the versed Sine of the Arc A C, and A L the versed Sine of the Arc AM BM-BA=BC-BA. The Ifosceles Triangles CFM, PON, are equiangular, since MF, NO, as also CF, PO (by 16 El. 11.) are parallel, Wherefore, if Perpendiculars CH PG, be drawn to the Sides FM, O N, the Triangles will be divided fimilarly, and we shall have FM: ON::MH:GN. Also, because the Triangles AO E, DIH, DLM,
are equiangular, we shall have A E: AO::IL: MH.
But it has been proved, that F M:ON::MH: GN. Wherefore it shall be, as A EXFM:A OX ON::ILXMH: MHxGN, or so is I L to GN; that is, the Rectangle under the Sines of the Legs, is to the Square of Radius, as the Difference of the verled Sines of the Base, and the Difference of the Legs BC, B A, is to the versed Sine of the Angle B. W. W.D.
PROPOSITION XXXVII. Tbe Difference of the versed Sines of two Arcs,
drawn into half the Radius, is equal to the Rectangle under the Sine of half the Sum, and the Sine of half the Difference of those Arcs.
LET there be two Arcs, BE, BF, whose Dif
ference E F let be bisected in D; then shall BD be the half Sum, and F D the half Difference of those Arcs GE=IL is the Difference of the versed Sines of the Arcs BE, BF; also, FO is the Sine of the half Difference of the Ares. And because the Triangles CDK, FEG, are equiangular, we have DK: GE:: (CD:FE::) C D E F E. Whence DKXF E, or D KXFO=G EXC D=ILX CD. "W.W.D.
PROPOSITION XXXVIII. Tbe versed Sine of any Arc, drawn into half the
Radius, is equal to the Square of the Sine of one half of the said Arc, TH HE Triangles C BM, DE B, are equiangular,
since the Angles at M and E are Right Angles, and the Angle at B is common.
Wherefore EB : BD::BM, :BC. And then will E BXBCS BM ~ BD; and E B X. BC=BM RBD= BMq. W. W.D.