PROP. XXVIII. The same construction remaining, the cosines of the segments of the vertical angle are reciprocally proportional to the tangents of the sides. Because (21.), cos BCD: R:: tan CD: tan BC, and also, cos ACD R tan CD: tan AC, by equality inversely, cos BCD: cos ACD tan AC: tan BC. Q. E. D. : :: : PROP. XXIX. If from an angle of a spherical triangle there be drawn a perpendicu lar to the opposite side, or base, the rectangle contained by the tan gents of half the sum, and of half the difference of the segments of the base is equal to the rectangle contained by the tangents of half the sum, and of half the difference of the two sides of the triangle. Let ABC be a spherical triangle, and let the arch CD be drawn from the angle C at right angles to the base AB, tan 1 (m+n) × tan (m-n) tan (a+b)x tan (a-b). = : Let BC-a, AC=b; DD=m, AD=n. Because (26.) cos a: cos b:: cosm: cos n, (E. 5.) cos a + b cos a-cos b:: cos m+cus n COS mcos n. But (1. Cor. 3. Pl. Trig.), cos a+cos b: cos a--cos b :: cot (a+b): tan(a-b), and also, cos m+cos n: cos m—cos n :: cot 1 (m+n): tan (m-n). Therefore, (11. 5.) cot } (a+b) : tan (ab): cot (m+n): tan (m-n). And because rectangles of the same altitude are as their bases, tan 1⁄2 (a + b) × cot (a+b): tan (a+b)× tan ↓ (∞ −b) :: tan (m + n) X cot i (m+n) : tan (mXn) +tan (m-n). Now the first and third terms of this proportion are equal, being each equal to the square of the radius, (1. Cor. Pl. Trig.), therefore the remaining two are equal, (9. 5.) or tan (mn) Xtan (m-n) = tan (a + b) Xtan (ab); that is, tan (BD+AD) X tan (BD-AD) tan (BC+ AC) x tan =tan × 1⁄2 (BC AC). Q. É. D. I COR. 1. Because the sides of equal rectangles are reciprocally proportional, tan (BD + AD) : tan (BC+AC) :: tan (BC-AC): tan (BD-AD). COR. 2. Since, when the perpendicular CD falls within the triangle, BD + AD AB, the base; and when CD falls without the triangle BD-AD AB, therefore in the first case, the proportion in the last corollary becomes, tan (AB): tan (BC+AC): : tan (BD-AD); and in the second case, it becomes by inversion and alternation, tan (AB): tan (BC+ AC) :: tan BC-AC) tan (BD+AD). (BC AC): tan THE preceding proposition, which is very useful in spherical trigoaometry, may be easily remembered from its analogy to the proposition in plane trigonometry, that the rectangle under half the sum, and half the difference of the sides of a plane triangle, is equal to the rectangle under half the sum, and half the difference of the segments of the base. See (K. 6.), also 4th Case Pl. Tr. We are indebted to NAPIER for this and the two following theorems, which are so well adapted to calculation by Logarithms, that they must be considered as three of the most valuable propositions in Trigonometry. PROP. XXX. If a perpendicular be drawn from an angle of a spherical triangle to the opposite side or base, the sine of the sum of the angles at the base is to the sine of their difference as the tangent of half the base to the tangent of half the difference of its segments, when the perpendicular falls within; but as the co-tangent of half the base to the co-tangent of half the sum of the segments, when the perpendicular falls without the triangle: And the sine of the sum of the two sides is to the sine of their difference as the co-tangent of half the angle contained by the sides, to the tangent of half the difference of the angles which the perpendicular makes with the same sides, when it falls within, or to the tangent of half the sum of these angles, when it falls without the triangle. If ABC be a spherical triangle, and AD a perpendicular to the base BC, sin (C+B): sin (C-B): tan BC: tan (BD-DC), when AD falls within the triangle; but sin (C+B): sin (CB); cot BC: cot (BD+DC), when AD falls without. And again, sin (AB+AC): sin (AB-AC) :: cot BAC : tan (BAD-CAD), when AD falls within; but when AD falls without the triangle, sin (AB+AC) sin (AB-AC :: cot BAC : tan (BAD+CAD). For in the triangle BAC (27.), tan B: tan C:: sin CD: sin BD, and therefore (E. 5.), tan C+ tan B: tan C-tan B :: sin BD + sin CD sin BD-sin CD. Now, (by the annexed Lemma) tan C + tan B tan C-tan B:: sin (C+B): sin (CB), and sin BD+ sin CD: sin BD-sin CD :: tan (BD+CD) : tan (BD-CD),(3. Pl. Trig.), therefore, because ratios which are equal to the same ratio are equal to one another (11. 5.), sin (C+B): sin (C-B) :: tan (BD+CD): tan (BD - CD). B D C B D Now when AD is within the triangle, BD+CD=BC, and therefore sin (C+B) sin (C- B) :: tan BC: tan (BD-CD). And again, when AD is without the triangle, BD-CD=BC, and therefore sin (C+B) sin (C.-B): : tan (BD+CD) : tan 1 BC, or because the tangents of any two arches are reciprocally as their co-tangents, sin (C+B) sin (CB) cot BC: cot 1 (BD+CD). The second part of the proposition is next to be demonstrated. Because (28.) tan AB tan AC cos CAD : cos BAD, tan AB + tan AC tan AB tan AC: cos CAD+cos BAD: cos CAD-cos BAD. But (Lemma) tan AB+tan AC: tan AB-tan AC :: sin (AB+AC): sin (AB-AC), and (1. cor. 3. Pl. Trig.) cos CAD + cos BAD: cos CAD-cosBAD: cot (BAD+CAD): tan (BAD-CAD). There fore (11. 5.) sin (AB+AC): sin (AB-AC) :: cot (BAD+CAD): tan (BAD-CAD). Now, when AD is within the triangle, BAD+CAD=BAC, and therefore sin (AB+AC): sin (AB-AC) : : cot BAC : tan (BAD-CAD). But if AD be without the triangle, BAD-CAD=BAC, and therefore sin (AB+AC): sin (AB-AC): : cot (BAD+CAD): tan BAC; or because cot (BAD+CAD): tan BAC:: 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) sin (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+B) cos A x cos B' sin (A+B) cos AX cos B sin (A-B) cos Ax cos B sin A cos A sin B + cos B =tan A, tan A+tan B In the same manner it is proved that tan A-tan B Therefore tan Atan 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 tangent of half the sum of the sides opposite. Let C+B=2S, C-B-2D, the base BC=2B, and the difference of the segments of the base, or BD-CD=2X. Then, because (30.) sin (C+B) sin (C-B): : tan BC: tan (BD - CD), sin 2S: sin 2D tan B: tan X. Now, sin 25 sin (S+S)=2 sin S x cos S, (Sect. III. cor. Pl. Tr.). In the same manner, sin 2D= 2 sin D x cos D. Therefore sin S x cos S: sin D x cos D :: tan B: tan X. by equals, tan ▲ X = tan B tan Σ But (29.) tan tan and therefore, = 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=2 sin (C+B)× cos (C-B), (Sect. III. 7. Pl. Tr.) 2 sin SXcos D; and sin C- sin B=2 cos(C+B) Xsin (C-B)= 2 cos SXsin D. Therefore 2 sin SX cos D: 2 cos S X sin D :: sin AB + sin AC : sin AB - sin AC. But (3. Pl. Tr. AB+sin AC sin AB-sin AC :: tan (AB+AC): tan (AB-AC):: tan : tan A, Σ being equal to (AB+ AC) and A to Therefore sin SXcos D: cos SXsin D :: tan : tan A cos SXsin D tan Σ sin SX cos D' (sin D) sin (AB-AC.). : tan A. Since then by multiplying equals X cos SXcos D____(sin D)3 = (sin S) X cos S x cos D (sin S)2 (BD-DC) tan (AB+AC) tan X tan X tan A tan B tan X tan A (sin D)3 But ; whence tan B sin D sin S' tan B tan (tan A)2__(sin D)3 (tan B)(sin S)2 tan X tan ▲ (tan 4). = ; and tan B or sin S (C-B) :: tan proposition. : BC : tan } (AB~AC); which is the first part of the sin D: : tan B: tan A, that is, sin (C+B): sin Again, since tan A cos SXsin D tan sin SXcos D. tan Σ sin SXcosD' or inversely tan X tan A cos SX sin D sin DXcos D and since tan B sin SXcos S ; therefore by multiplication,, tan X X 'tan B |