now suppose the spherical angle A to be right, we have (equa. 5, chap. vi) sin b = sin a sin B. But it has been proved (chap. iv. equa. y) that the sine of any arc a is equal to the sum of the following series:

And, in the present inquiry, all the terms after the second may be neglected, because the 5th power of an arc of 4° divided by 120, gives a quotient not exceeding 0'.01. Consequently, we may assume sin b = b — 1b3, sin a = a - a3; and thus the preceding equation will

become

Now, if the triangle were considered as rectilinear, we should have b= a sin B; a theorem which manifestly gives the side b or AC too great by (a3 sin B 63). But, neglecting quantities of the fifth order, for the reason already assigned, the last equation but one gives 63 = a3 sin3 B. Therefore, by substitution, e = — sin B (1 sin B): or, to have this error in seconds, take R the radius expressed in seconds, so shall

Cor. 1. If a = 4°, and в 35° 16′, in which case the value of sin в cos в is a maximum, we shall find - 41′′.

e=

Cor. 2. If, with the same data, the correction be applied, to find the side c adjacent to the given angle, we should have

So that this error exists in a contrary sense to the other; the one being subtractive, the other additive.

Cor. 3. The data being the same, if we have to find the angle c, the error to be corrected will be

As to the excess of the arc over its chord, it is easy to find it correctly from the expressions in prob. 8: but for arcs that are very small, compared with the radius, a near approximation to that excess will be found in the same measures as the radius of the earth, by taking of the quotient of the cube of the length of the arc divided by the square of the radius.

PROBLEM XI.

It is required to investigate a theorem, by means of which, spherical triangles, whose sides are small compared with the radius, may be solved by the rules for plane trigonometry, without considering the chords of the respective arcs or sides.

The sum of the angles of a spherical triangle exceeds 180°, by a quantity E called (chap. vi. p. 102) the spherical excess, which' is a correct measure of the area of the triangle.

The sum of the angles of a spherical triangle ABC, exceeds 180° by 2r2. AC. AB sin A = = surface of the triangle; r being the

sphere. Or, when r =

1, we have

AB. Ac sin a

A + B + C = 180° +

r2AB. AC sin A

radius of the

= 180° + LAB. Ac sin A + AB. CB sin B + AC. BC sin c..

Distributing this excess equally, by taking its third part from each angle, we shall have

who have carefully studied chapters iv. and vi. of this work:

sin A

Bcsin B

sin A

1

cot B tan E

cot A tan E

(1 + tan E Cot A tan E cot B)

-

(1 + AB. AC sin a cot A➡ AB. BC

sin B cot B)

BC sin B

sin A

BC sin B

sin A

(1 +

AB. AC COS A —

AR.BC COS B)

(1 + AB. AD — LAB, BD)

CAD and BD being segments of the base AB by a perpendicular CD from c]

Hence, ACÁC3 = (BC — ÷BC3)

But (chap. iv. equa. Y) sin AC and sin BC BC- BC3, nearly; prob. 10, being imperceptible in

quently,

sin AC =

sin B

sin A

AC - AC3, nearly, the error, as shown small arcs.

sin BC sin B

sin A

Conse

And thus we have deduced the truth of the following theorem, viz..

A spherical triangle being proposed, of which the sides are very small, compared with the radius of the sphere; if from each of its angles one third of the excess of the sum of its three angles above two right angles be subtracted, the angles so diminished may be taken for the angles of a rectilinear triangle, whose sides are equal in length to those of the proposed spherical triangle.*

SCHOLIUM.

We have already given (in sect. 6, chap. vi.) expressions for finding the spherical excess. A few additional rules may with propriety be presented here.

1. The spherical excess E, may be found in seconds,

by the expression, E = where s is the surface of the triangle (whose sides are a, b, c, and angles A, B, C)

= 1bc sin A = lab sin c = Jac sin B‚— ba?

sin, B-sin c

sin (+c)'

is the radius of the earth, in the same measures as a, b, and c, and R" = 206264" 8, the seconds in an arc equal in length to the radius.

If this formula be applied logarithmically, then log Blog5:3144251,

บ

are

This curious theorem was first announced by M. Legendre, in the Memoirs of the Paris Academy, for 1787. The investiga tion here given is by M. Delambre.

2. From the logarithm of the area of the triangle, taken as a plane one, in feet, subtract the constant log 9.3267737, then the remainder is the logarithm of the excess above 180°, in seconds nearly.*

R"

2r2

3. Since sbc sin A, we shall manifestly have E =

bc sin A. Hence, if from the vertical angle в we demit the perpendicular BD upon the base AC, dividing it into two segments «, B, we shall havé b « + 6, and

sin A.

2,2

But the two right angled triangles ABD, CBD, being nearly rectilinear, give a a cos c, and 3 = c cos A; whence we have

E

R"

2r2

R" 2r2

ac sin A cos C+ c2 sin A cos A

In like manner, the triangle ABC, which itself is so small as to differ but little from a plane triangle, gives c sin a = a sin c. Also, sin a cos A = sin 2A, and sin c cos c = sin 2c (equa. 1, chap. iv). Therefore, finally,

1=3

From this theorem a table may be formed, from which the spherical excess may be found; entering the table with each of the sides above the base and its adjacent angle, as arguments.

4. If the base b and height h, of the triangle are given, then we have evidently e3bh Hence results the

=

R"

following simple logarithmic rule: add the logarithm of the base of the triangle, taken in feet, to the logarithm of the perpendicular, taken in the same measure; deduct from the sum the logarithm 9-6278037; the re

This is Mr. Dalby's rule given by General Roy in the Philosophical Transactions, for 1790, p. 171.

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