Case 1. Let us suppose the three points A, P, E, fig. 15, plate 9, to be equally distant from the centre of the earth, and that the point R is higher than these points by the distance or quantity R E; now it is required to reduce the triangle APR to that of A PE.

By the following rule, you may reduce the angles RAP, RPA, which have their summits in the plain of reduction, to the angles EPA, EAP.

Rule. The cosine of the reduced angle is equal to the cosine of the observed angle, divided by the cosine of the angle of elevation.

These two angles being known, the third E is consequently known; we shall, however, give a rule for finding A EP, independent of the other

two.

Rule. The cosine of the reduced angle is cqual to the cosine of the observed angle, lessened by the rectangle of the sines of the angles of elevation, divided by the rectangle of the cosine of the same angles.

The reduction of the sides can be no difficulty.

Case 2. Fig. 17, plate 9. Let A Rr, be the triangle to be reduced to the plain A Ee, the points E, e, of the vertical lines RE, re, being supposed equally distant from the centre of the earth.

Prolong the plain A Ee, to P, that is, till it meets the line Rr, produced to P; and the value of EA e, will be found by the following formula.

1. Tangent (PAR+PAr) = tangent + RArx tangent + (RAE+rAe). Knowing the half sum tangent (RAE-rAe)

and half difference of P Á R, and PA r, we obtain the value of each of the angles; the value of P AE, and PA e, may be then obtained by the first of the two preceding rules, and the difference between them is the angle sought.

Let C, fig. 16, plate 9, be the centre of the earth, let AB, be the side of a triangle reduced to

a common horizon by the preceding methods; if it be required to reduce this to the plain DE, as these planes are parallel, the angles will remain the same; therefore, the sides only are to be reduced, the mode of performing which is evident from the figure.

Method of referring a series of triangles to a meridian line, and another line perpendicular to it. This method will be found somewhat similar to one used by Mr. Gale, and described at length in the article of surveying; it is a mode that should be adopted wherever extreme accuracy is required, for whatever care is taken to protract a series of triangles, the protractor, the points of the compasses, the thickness of the line, the inequality of the paper, &c. will produce in the fixing of the points of a triangle an error, which, though small at first, will have its influence on those that succeed, and become very sensible, in proportion as the number of triangles is augmented. This multiplication of errors is avoided by the following problem.

Let AB, fig. 14, plate 9, be the meridian, CD, the perpendicular, and the triangles oad, dae, deg, egi, gil, those that have been observed; from the point o, (which is always supposed to be on a meridian, or whose relation to a meridian is known) observe the angle Boa, to know how much the point a declines from the meridian.

In the right-angled triangle, o Ba, we have the angle Boa, and the right angle, and consequently, the angle oa B, together with the side o a, to find o B, and Ba.

For the point d, add the angle Boa, to the observed angle a o d, for the do b, or its equal o d m, and the complement is the angle mod, whence as before, to find om, and m d.

For the point g, add the angles m do, oda, a de, and edg, which subtract from 360, to obtain the angle gdr, of the right angled triangle grd; hence

Case 1. Let us suppose the three points A, P, E, fig. 15, plate 9, to be equally distant from the centre of the earth, and that the point R is higher than these points by the distance or quantity R E; now it is required to reduce the triangle APR to that of A PE.

By the following rule, you may reduce the angles RAP, RPA, which have their summits in the plain of reduction, to the angles EPA, EAP.

Rule. The cosine of the reduced angle is equal to the cosine of the observed angle, divided by the cosine of the angle of elevation.

These two angles being known, the third E is consequently known; we shall, however, give a rule for finding AEP, independent of the other two.

Rule. The cosine of the reduced angle is equal to the cosine of the observed angle, lessened by the rectangle of the sines of the angles of elevation, divided by the rectangle of the cosine of the same angles.

The reduction of the sides can be no difficulty.

Case 2. Fig. 17, plate 9. Let A Rr, be the triangle to be reduced to the plain A Ee, the points E, e, of the vertical lines RE, re, being supposed equally distant from the centre of the earth.

Prolong the plain A Ee, to P, that is, till it meets the line Rr, produced to P; and the value of E A e, will be found by the following formulæ.

1. Tangent (PAR+PAr) = tangent & R Ar× tangent (RAE+rAe). tangent (RAE-rAe)

Knowing the half sum

and half difference of P A R, and P Ar, we obtain, the value of each of the angles; the value of PA E, and PA e, may be then obtained by the first of the two preceding rules, and the difference between them is the angle sought.

Let C, fig. 16, plate 9, be the centre of the earth, let AB, be the side of a triangle reduced to

a common horizon by the preceding methods; if it be required to reduce this to the plain DE, as these planes are parallel, the angles will remain the same; therefore, the sides only are to be reduced, the mode of performing which is evident from the figure.

Method of referring a series of triangles to a meridian line, and another line perpendicular to it. This method will be found somewhat similar to one used by Mr. Gale, and described at length in the article of surveying; it is a mode that should be adopted wherever extreme accuracy is required, for whatever care is taken to protract a series of triangles, the protractor, the points of the compasses, the thickness of the line, the inequality of the paper, &c. will produce in the fixing of the points of a triangle an error, which, though small at first, will have its influence on those that succeed, and become very sensible, in proportion as the number of triangles is augmented. This multiplication of errors is avoided by the following problem.

Let A B, fig. 14, plate 9, be the meridian, CD, the perpendicular, and the triangles o ad, dae, deg, egi, gil, those that have been observed; from the point o, (which is always supposed to be on a meridian, or whose relation to a meridian is known) observe the angle Boa, to know how much the point a declines from the meridian.

In the right-angled triangle, o B a, we have the angle Boa, and the right angle, and consequently, the angle oa B, together with the side o a, to find o B, and Ba.

For the point d, add the angle Boa, to the observed angle a o d, for the do b, or its equal o d m, and the complement is the angle mod, whence as before, to find om, and m d.

For the point g, add the angles m d o, oda, ade, and edg, which subtract from 360, to obtain the angle gdr, of the right angled triangle grd; hence

Case 1. Let us suppose the three points A, P, E, fig. 15, plate 9, to be equally distant from the centre of the earth, and that the point R is higher than these points by the distance or quantity R E; now it is required to reduce the triangle APR to that of A PÉ.

By the following rule, you may reduce the angles RAP, RPA, which have their summits in the plain of reduction, to the angles EPA, E AP.

Rule. The cosine of the reduced angle is equal to the cosine of the observed angle, divided by the cosine of the angle of elevation.

These two angles being known, the third E is consequently known; we shall, however, give a rule for finding AEP, independent of the other two.

Rule. The cosine of the reduced angle is equal to the cosine of the observed angle, lessened by the rectangle of the sines of the angles of elevation, divided by the rectangle of the cosine of the same angles.

The reduction of the sides can be no difficulty. Case 2. Fig. 17, plate 9. Let A R r, be the triangle to be reduced to the plain A Ee, the points E, e, of the vertical lines RE, re, being supposed equally distant from the centre of the earth.

Prolong the plain A E e, to P, that is, till it meets the line Rr, produced to P; and the value of E A e, will be found by the following formula.

1. Tangent (PAR+PAr) = tangent + R Ar× tangent + (RAE+rAe). Knowing the half sum tangent (RAE-rAe)

and half difference of P A R, and PA r, we obtain, the value of each of the angles; the value of PA E, and PA e, may be then obtained by the first of the two preceding rules, and the difference between them is the angle sought.

Let C, fig. 16, plate 9, be the centre of the rth, let AB, be the side of a triangle reduced to