PROP. XXIII.

In right angled spherical triangles, the cosine of either of the sides is to the radius, as the cosine of the angle opposite to that side is to the sine of the other angle.

The same construction remaining: In the triangle CEF, sin. CF R:: sin. EF: sin. ECF, (19.); but sin. CF= =cos. CA, sin. EF=cos. ABC and sin. ECF=sin. BCA; therefore, cos: CA: R:: cos. ABC: sin. BCA. Q. E. D.

PROP. XXIV.

In spherical triangles, whether right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them.

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First, Let ABC be a right angled triangle, having a right angle at A; therefore, (19.) the sine of the hypotenuse BC is to the radius, (or the sine of the right angle at A), as the sine of the side AC to the sine of the angle B. And, in like manner, the sine of BC is to the sine of the angle A, as the sine of AB to the sine of the angle C; wherefore (11. 5.) the sine of the side AC is to the sine of the angle A B, as the sine of AB to the sine of the angle C.

Secondly, Let ABC be an oblique angled triangle, the sine of any of the sides BC will be to the sine of any of the other two AC, as the sine of the angle A opposite to BC, is to the sine of the angle B op2 posite to AC. Through the point C, let there be drawn an arch of a great circle CD perpendicular to AB; and in the right angled trian

gle BCD, sin. BC: R:: sin. CD: sin. B, (19.); and in the triangle ADC, sin. AC: R: sin. CD: sin. A; wherefore, by equality inversely, sin. BC sin. AC :: sin. A : sin. B. In the same manner, it may be proved that sin. BC: sin. AB :: sin. A: sin. C, &c. Therefore, &o. Q. E. D.

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PROP. XXV.

In oblique angled spherical triangles, a perpendicular arch being drawn from any of the angles upon the opposite side, the cosines of the angles at the base are proportional to the sines of the segments of the vertical angle.

Let ABC be a triangle, and the arch CD perpendicular to the base BA; the cosine of the angle B will be to the cosine of the angle A, as the sine of the angle BCD to the sine of the angle ACD.

For having drawn CD perpendicular to AB, in the right angled triangle BCD, (23.) cos. CD: R:: cos. B: sin. DCB; and in the right angled triangle ACD, cos. CD: R:: cos. A: sin. ACD; therefore (11. 5.) cos. B: sin. DCB :: cos. A: sin. ACD, and alternately, cos: B: cos. A :: sin. BCD : sin. ACD. Q. E. D.

PROP. XXVÍ.

The same things remaining, the cosines of the sides BC, CA, are proportional to the cosines of BD, DA, the segments of the base.

For in the triangle BCD, (22.), cos. BC: cos. BD:: cos. DC : R, and in the triangle ACD, cos. AC: cos. AD :: cos. DC: R; therefore (11. 5.) cos. BC: cos. BD:: cos. AC: cos. AD, and alternately, cos. BC: cos. AC: cos. BD: cos. AD. Q. E. D.

PROP. XXVII.

The same construction remaining, the sines of BD, DA the segments of the base are reciprocally proportional to the tangents of B and A, the angles at the base.

In the triangle BCD, (18.), sin. BD: R:: tau. DC: tan. B; and in the triangle ACD, sin. AD: R:: tan. DC: tau. A; therefore, by equality inversely, sin. BI): sin. AD :: tan. A : tan. B. Q. E. D.

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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 perpendicular to the opposite side, or base, the rectangle contained by the tangents 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. (m+n) xtan. (m-n) tan. (a+b)× tan. (a-b).

Let BC=a, AC=b; DD=m, AD=n. Because (26.) cos. a : cos. b:: cos. m : cos. n, (E. 5.) cos. a+b: cos, a—cos. b :: cos.m+cos. n: cos.m-cos. n. But (1. Cor. 3. Pl. Tr.), cos. a+cos. b: cos. a➡ cos. b

cot. (a+b): tan. (a−b,) and also, cos. m+cos. n: cos.m-cos. ncot. (m+n): tan. (m-n). Therefore, (11. 5.) cot. (a+b); tan. (a-b) cot. (m+n): tan. (m-n). And because rectangles of the same altitude are as their bases, tan. (a+b) × cot. § (a+b): tan. (a+b)xtan. (a-b) tan. (m+n) xcot. (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. Tr.), therefore the remaining two are equal, (9. 5.) or tan. (m+n)xtan. (m-n)=tan. § (a+b)×tau. (ab); that is, tan. (BD+AD)xtan. (BD-AD)=tan. (BC+AC)xtan. (BC-AC). Q. É. D.

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. (BC-AC): tan. (BD-AD); and in the second case, it becomes by inversion and alteration, tan. (AB): tan. (BC+AC) ;; tan. (BC−AC): tan. (BD+AD).

The preceding proposition, which is very useful in spherical trigonometry, 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 seg ments 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. (C—B) :: cot. BC ent. (BD+DC), when AD falls without. And again,

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sin. (AB+AC): sin. (AB-AC) :: cot. BAC: tan. 1⁄2 (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. (C−B), and sin. BD-+sin. CD sin. BD-sin. CD tan. (BD+CD): tan. (BD-CD), (3. Pl. Tr.), therefore, because ratios which are equal to the same ratio are equal to one another (11. 5.), sin. (C+B) : sin. (C—B) :: țan. § (BD+CD): tan. (BD-CD).

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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. BC, or because the tangents of any two arches are reciprocally as their co-tangents, sin. (C+B): sin. (C-B) cot. BC cot. (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. eor. 3. Pl. Tr.) cos. CAD+cos. BAD: cos. CAD-cos, BAD: cot. & (BAD+CAD) : tan. § (BAD -CAD). Therefore (11.5.) sin. (AB+AC): sin. (AB-AC): cot. (BAD+CAD): tan. 1⁄2 (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.)