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:: tan DC: tan B; and in the triangle ACD, sin AD: R::tan DC: tan A ; therefore, by equality inversely, sin BD: sin AD:: tan A: tan B.. Q. E. D.

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. QË.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 Cat right angles to the base AB; tan (BD+AD) x tan (BD-AD) =tan (BC+ AC) x tan (BC—AC).

Let BC a, AC = 6; BD = m, AD = n. Because (26.) cos a: cos b:: cos m : cos n, (E. 5.) cos a + cos b: : cos m + cos n: cos m

cos a

cos m

cos b

cos 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 n :: cot (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) x cot (a+b): tan (a + b) x tan (a—b) :: tan (m + n) x cot (m + n): tan (m + n) × 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 (m + n) x tan (m-n) tan (a+b) x tan (a+b); that is, tan(BD + AD) × tan (BD-AD) = tan (BC + AC) × tan § (BC—AC). Q. E. D.

COR, 1. Because the sides of equal rectangles are reciprocally proportional, tan (BD + AD) : tan ¦ (BC + AC) : : tan 1⁄2 (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 alternation, tan ↓ (AB) : tan } (BC + AC) :: tan (BC-AC) : tan (BD + AD).·

D.A.

B

B

SCHOLIUM.

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 segments of the base. See (K. 6.), also 4th Case Pl. Trig. 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 diffe rence of its segments, when the perpendicular falls within; but as the tangent of half the base to the tangent of half the sum of the segments, when the perpendicular falls with out 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 tri

B

D

angle; but sin (C + B): sin (C-B): : tan BC tan (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 (ABAC) cot BAC: tan (BAD + CAD).

For, in the triangle BAC, when AD is within the triangle, tan B: tan C:: sin CD: sin BD (27.), and therefore, (E. 5.), tan C + tan B: tan C-tan B:: sin BD + sin CD: sin BD-sin CD. Now, (by the annexed Lem

ma) tan C+tan B: tan C-tan B: sin (C4B): sin (CB), and sin BD + sin CD: sin BD-sin CD: ; tan ¦ (BD + CD) : tan} (BD-CD), (3. Pl. Trig.), there fore, 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).

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'R

AD

:

B

If.

D

(180 the

::

Again, when AD is without the triangle, because tan Btan ACD:: sin CD sin BD; that is, angle ACB by the letter C), tan B: tan sin CD sin BD; therefore, tan (180°-C)+tan B: tan (180°-C)-tan B:: sin BD + sin CD : sin BDsin CD. Now, tan (180°—C) + tan B: tan (180°C). tan B:: sin {180°—(C—B)} ; sin' {180°—(C + BY } (by the Lemma), or because the sine of an angle is the same as the sine of its supplement, (Pl. Trig. Def. 4.), tan (180°C) +tan B tan (180-C)-tan B: sin (C-B): sin (C+B), and sin BD + sin CD: sin BD -sin CD tan (BD + CD); tan (BD-CD); therefore, sin (C-B): sin (C + B) :: tan! (BD + CD) :tan (BD-CD); and, by inversion, and observing that BD-CD = BC, sin (C + B): sin (CB) : : tan BC: tan (BD + CD). AIT

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-cos BAD: : cot (BAD +CAD) : tan (BAD-CAD). Therefore (11.5.) sin (AB + AC):