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PROBLEM VII.

To reduce angles measured in a plane inclined to the horizon, to the corresponding angles in the horizontal plane.

Let BCA be an angle measured in a plane inclined to the horizon, and let B'CA' be the corresponding angle in the horizontal plane. Let d and d be the zenith distances, or the complements of the angles of elevation ACA', BCB'. Then from z the zenith of the observer, or of the angle c, draw the arcs za, zb, of vertical circles, measuring the zenith distances d, d', and draw the arc ab of another great circle to measure the angle c. It follows

d, zb

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d', ab

A

'A

= c,

from this construction, that the angle z, of the spherical triangle zab, is equal to the horizontal angle A'c'B; and that, to find it, the three sides za = are given. Call the sum of these s; then the corresponding formula of ch. vi. pa. 90, applied to the present instance, becomes

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If h and h' represent the angles of altitude ACA', BCB', the preceding expression will become

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Or, in logarithms,

log sin c(20+ log sin (c+hh) + log sin } (c + h − h) log cos hlog cos h').

Cor. 1. If hh', then is sin c =

sin ACB

cos h

; and

log sin A'CB10+ log sin ACB log cos h.

Cor. 2. If the angles h and h' be very small, and nearly equal; then, since the cosines of small angles

vary extremely slowly, we may, without sensible error, take

log sin A'CB' = 10 + log sin ACB log cos § (h + h'). Cor. 3. In this case the correction x = A'CB' — acb, may be found by the expression

d'

x = sin 1′′ (tan }c (10-d+d) 2

2

— cot fc (d=d')2).

And in this formula, as well as the first given for sin c, d and d' may be either one or both greater or less than a quadrant; that is, the equations will obtain whether ACA' and BCB' be each an elevation or a depression.

Scholium. By means of this problem, if the altitude of a hill be found barometrically, according to one of the methods described in Gregory's Mechanics, vol. i. book 5, or geometrically, according to some of those described in heights and distances, (chap. v.); then, finding the angles formed at the place of observation, by any objects in the country below, and their respective angles of depression, their horizontal angles, and thence their distances, may be found, and their relative places fixed in a map of the country; taking care to have a sufficient number of angles between intersecting lines, to verify the operations.

PROBLEM VIII.

In any spherical triangle, knowing two sides and the included angle; it is required to find the angle comprehended by the chords of those two sides.

A

b

Let the angles of the spherical triangle be A, B, C, the corresponding angles included by the chords A', B', C'; the spherical sides opposite the former, a, b, c; the chords respectively opposite the latter, a, 6, y; then, there are given b, c, and A, to find a ́.

B

a

C

C

a

Here, from equa. 2, chap. vi. we have

cos a sin b sin c cos A+ cos b cos c. But cos c = cos (c + c) = cos2 c - sin? ¿c (by art. 23, chap. iv.) sin2 c) sin2 c = 1 2

sinc.

= (1

And in like manner cos a = 1

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2 sin2a, and

cos b = 1 2 sin2 6. Therefore the preceding equa

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tion becomes

1 - 2 sin2 a 4 sin b cos 6 sin c cos c cos a +

(1-2 sin2 16) (1

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2 sin2 c).

ja, sin } = }, sản Ęc

=

= : which values

substituted in the equation, we obtain, after a little re

duction,

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But sin a =

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By cos A' By cos b cos e cos A + 1622; whence, dividing by By, there results

cos A'cos cos ac cos a + 13tv; or, lastly, by restoring the values of,, we have COS A' cos cos e cos a + sin 16 sin c. . . . (1.) Cor. 1. It follows evidently from this formula, that when the spherical angle is right or obtuse, it is always greater than the corresponding angle of the chords. Cor. 2. The spherical angle, if acute, is less than the corresponding angle of the chords, when we have coś A sin b sin c greater than

1.

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Knowing two sides and the included angle of a rectilinear triangle, it is required to find the spherical angle of the two arcs of which those two sides are the chords. Here, y, and the angle A' are given, to find a. Now, since in all cases, cos = √(1 − sin2), we have cos 6 cosc √[(1 sin2) (1-sin2 c)]; we have also, as above, sin be, and sin cfr.

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Substituting these values in the equation 1 of the preceding problem, there will result, by reduction,,

COS A =

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(2.) √(1 − }6). (1 + 16) ̃ ̈. (1 − §2) . (1 + y) To compute by this formula, the values of the sides, , must be reduced to the corresponding values of the chords of a circle whose radius is unity. This is easily effected by dividing the values of the sides given in-feet, or toises, &c. by such a power of 10, that neither of the sides shall exceed 2, the value of the greatest chord, when radius is equal to unity.

From this investigation, and that of the preceding problem, the following corollaries may be drawn.

Cor. 1. If c= b, and of consequence y = 6, then will cos A' = cos a cosc + sin? c; and thence 1-2 sin A' (1-2 sin2 A) cos2c+ (1 cos2 (c): from which may be deduced

=

sin A sina cos c. (3.):

Cor. 2. Also, since.cos c

....

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✓(1 sinc)=√(L

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- y2), equa. 2 will, in this case, reduce to

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sin A'

sin A=

√(1 − §y). (1 + έy)

(4.)

Cor. 3. From the equa. 3, it appears that the vertical angle of an isosceles spherical triangle, is always greater than the corresponding angle of the chords,

Cor. 4. If A 90°, the formula 1, 2, give

=

(5.)

COS A' = sin b sin cry. = These five formulae are strict and rigorous, whatever be the magnitude of the triangle. But if the triangles be small, the arcs may be put instead of the sines in equa. 5, then

Cor. 5. As cos A' = sin (90°· A) in this case, 90° A; the small excess of the spherical right angle over the corresponding rectilinear angle, will, supposing the arcs b, c, taken in seconds, be given in seconds by the following expression:

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The error in this formula will not amount to a second, when bc is less than 10°, or than 700 miles measured on the earth's surface.

Cor. 6. If the hypothenuse does not exceed 110, we may substitute a sin c instead of c, and a cos c instead of b; this will give bc a sin c cos c = a2 sin 2 (90°

B) a2 sin 28: whence

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If a = 110, and B = c = 45° nearly; then will 90° — A'

= 177.

Cor. 7. Retaining the same hypothesis of A 90°, and a = or < 11°, we have

B

BB

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Also c➡ ➡

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Cor. 8. Comparing formula 8, 9, with 6, we have B = C C = (90° A). Whence it appears that the sum of the two excesses of the oblique spherical angles, over the corresponding angles of the chords, in a small right angled triangle, is equal to the excess of the right angle over the corresponding angle of the chords. So that either of the formulæ 6, 7, 8, 9, will suffice to determine the difference of each of the three angles of a small right angled spherical triangle, from the corresponding angles of the chords. And hence this method may be applied to the measuring an are of the meridian by means of a series of triangles.

PROBLEM X.

In a spherical triangle ABC, right angled in A, knowing the hypothenuse BC (less than 4°) and the angle B, it is required to find the error e committed through finding by plane trigonometry, the opposite side Ac.

Referring still to the diagram of prob. 8, where we

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