sın (AB+AC): sin (AB-AC) :: co when AD falls within: but when AD sin (AB+AC): sin (AB-AC) :: cot BAC: tan (BAD-CAD), falls without the triangle, ta BAC: tan (BAD-CAD). Men C:: sin CD: sin BD, I tan B:: sin BD+sin ALemma) tan C+tan Fo For in the triangle BAC (27.), tan B: and therefore (E. 5.), tan C+tan B: tan CCD: sin BD-sin CD. Now, (by the annexed B: tan C-tan B :: sin (C+B): sin (C-B), and sin BD - sin CD :: tan(BD+CD): tan(BD-CD), (3. Pl.Trig therefore, because ratios which are equal to the san to one another (11. 5.), sin (C+B): sin (C-B) :: tabDb tae ratio are equal tan (BD-CD). COS sin COS B Now when AD is within the triangle, BD+CD=BC, and therefore sin (C+B): sin (C-B)::tan BC: tan (BD-CD). An and again when AD is without the triangle, BD-CD=BC, and therefore sin (C+B): sin (C-B) :: tan (BD+CD): tan BC, or because the tangents of any two arches are reciprocally as their co-tangents, in (C+B): sin (C-B):: cot BC: cot (BD+CD). the di th en ta Le of the The second part of the proposition is next to be demonstrated. Because (28.) tan AB: tan AC:: cos CAD: cos BAD, tan AL+C AC: tan AB-tan AC::cos CAD+cos BAD : : COS CAD-Cos But (Lemma) tan AB+tan AC: tan AB-tan AC :: sin (AB+ AC S sin (AB-AC), and (1. cor. 3. Pl. Trig.) cos CAD+cos BAL CAD-cosBAD::cot (BAD+CAD): tan (BAD-CAD). T fore (11.5.) sin (AB+AC): sin (AB-AC):: cot (BAD+CA tan (BAD-CAD). Now, when AD is within the triangle, B +CAD=BAC, and therefore sin (AB+AC): sin (AB-AC) :: BAC: tan (BAD-CAD). D But if AD be without the triangle, BAD-CAD=BAC, and there fore sin (AB+AC): sin (AB-AC) :: cot (BAD+CAD): tan BAC; or because cot (BAD+CAD): tanBAC:: cot BAC: tan (BAD+CAD), sin (AB+AC): sin (AB-AC) :: cot BAC: tan (BAD+CAD). Wherefore, &c. Q. E. D. LEMMA. The sum of the tangents of any two arches, is to the difference of their tangents, as the sine of the sum of the arches, to the sine of their difference. Let A and B be two arches, tan A+tan B: tan A-tan B :: sin (A+B): (A-B). For, by § 6. page 243, sin Axcos B+cos Axsin B=sin (A+B), and therefore dividing all by cos A cos B, sin A sin B + cos A cos B sin (A + B), that is, because. sin A =tan A, tan A+tan B cos Axcos B' cos A sin (A + B). In the same manner it is proved that tan A-tan B cos Axcos B sin (A - B) cos Ax cos B Therefore tan A + tan B: tan A-tan B :: sin A+B): sin (A-B). Q. E. D. PROP. XXXI. The sine of half the sum of any two angles of a spherical triangle is to the sine of half their difference, as the tangent of half the side adjacent to these angles is to the tangent of half the difference of the sides opposite to them; and the cosine of half the sum of the same angles is to the cosine of half their difference, as the tangent of half the side adjacent to them, to the angent of half the sum of the sides opposite. et C+B=2S, C-B=2D, the base BC=2B, and the difference le segments of the base, or BD-CD=2X. Then, because (30.) C+B): sin (C-B):: tan BC: tan (BD-CD), sin 2S: sin Stan B: tan X. Now, sin 2S=sin (S + S)=2 sin S x cos S, + III. cor. Pl. Tr.). In the same manner, sin 2D=2 sin D x AL Therefore sin S x cos S: sin D x cos D :: tan B:tan X. T , B. C):: Again, in the spherical triangle ABC it has been proved, that sin C+sin B: sin C-sin B :: sin AB+sin AC: sin AB-sin AC, and since sin C+sin B=2sin(C+B)+cos(C-B), (Sect. III. 7. Pl. Tr.)=2 sin Sxcos D; and sin C-sin B=2 cos (C+B)×sin (C-B)=2 cos Sxsin D. Therefore 2 sin SX cos D: 2 cos Sx sin D :: sin AB+sin AC: sin AB-sin AC. But (3. Pl. Tr.) sin AB+sin AC: sin AB-sin AC:: tan (AB+AC): tan (AB-AC):: tan 2 : tan ∆, 2 being equal to (AB+AC) and to (AB-AC.). Therefore sin Sxcos D: cos Sxsin D :: tan 2: tan ∆. Since then tan Xsin Dxcos D tan A cos Sxsin D = tan B sin Sxcos S' quals by equals, But ;and. tansin Sxcos D' by multiplying e tan X tan_(sin D) 2xcos Sxcos D_(sin D) tan Batan (sin S) xcos Sxcos D (sin S)** tan (AB+AC) , that is, tan BC (29.)tan(BD-DC) and therefore, tan B tan Btan Σ X (tan B)2 , as also tan tan X tan Σ tan B' tan X tan tan2 tan B tan Σ (tan B) (sin D) 2 2; whence (tan)_(sin D) 2 (tan B) (sin S); and 2 tan A tan B sin D , or sin S: sin D :: tan B: tan, that is, sin (C+B): sin sin S' (C-B):: tan BC: tan (AB-AC); which is the first part of the tan A cos Sxsin D tan Σ sin Sxcos D' and since tan X_sin Dxcos D tan B Sin Sxcos ; therefore by multiplication, 1 tan X tan X tan X_tan Extan, wherefore also tan B (tan B) 2 5, as has just been shewn. Therefore (cos D)_(tan 2), and consequently cos D_tan 2 (cos S) (tan B) 2' or cos cos Stan B S: cos. D:: tan. B: tan. 2, that is, cos. (C+B): cos. (C-B) :: tan. Cor. 1. By applying this proposition to the triangle supplemental to ABC (11.5.), and by considering, that the sine of half the sum or half the difference of the supplements of two arches, is the same with the sine of half the sum or half the difference of the arches themselves: and that the same is true of the cosines, and of the tangents of half the sum or half the difference of the supplements of two arches: but that the tangent of half the supplement of an arch is the same with the cotangent of half the arch itself; it will follow, that the sine of half the sum of any two sides of a spherical triangle, is to the sine of half their difference as the cotangent of half the angle contained between them, to the tangent of half the difference of the angles opposite to them: and also that the cosine of half the sum of these sides, is to the cosine of half their difference, as the cotangent of half the angle contained between them, to the tangent of half the sum of the angles opposite to them. COR. 2. If therefore A, B, C be the three angles of a spherical triangle, a, b, c the sides opposite to them, L. sin(A+B): sin (A-B):: tanc:tan (a−b). III. sin (a+b): sin(a−b)::tan C: tan IV. cos (a+b): cos (a−b)::tan C: tan (A-B). (A+B). PROBLEM I. In a right angled spherical triangle, of the three sides and three angles, any two being given, besides the right angle, to find the other three. This problem has sixteen cases, the solutions of which are contained in the following table, where ABC is any spherical triangle right angled at A. BC and B. 1 AC. R: sin BC : : sin B: sin AC, (19). C. |