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.

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), e sine as the sme of the fo

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 OT
AB to the sine of the angle
C; wherefore (11. 5.) the.
sine of the side AC is to

the

A.

sine of the angle 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 opposite 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

C

AD

B

B

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, C. Q.E. D.

PROP. XXV.

In oblique angled spherical triangles, a perpendi cular 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 perpendicu lar 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 Asin therefore (11. 5.)

cos B: sin DCB:: cos A: cos B cos A :: sin BCD : sin

and alternately, Q. E. D.

PROP. XXVI.

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:: 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, 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.

cos b::

Let ABC be a spherical triangle, and let the arch CD be drawn from the angle Cat right angles to the base AB, tan (m+n) x tan (m-n) tan (a+b) x tan (a—b). Let BC-a, AC6; DD m, AD = n. Because (26.) cos a: cos b:: cos m : cos n, (E. 5.) cos a + cos b: cos a-cos b:: cos m + cos n: cos m cos n. But (1. Cor. 3, Pl. Trig.), cos a + cos b: cos a cot (a+b) tan (a - b), and also, cos m + cos n: cos m 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 1 (a—b): : tan ÷ (m+n) × cot(m+n): tan (m+n) x 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 (mn) = tan (a + b) x tan (ab); 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 (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 în 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 1 (BD+AD).

C

B

A

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. 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.