CHAPTER IV.

MISCELLANEOUS APPLICATIONS.

124. To find the distance between two points on the earth's surface and their direction from each other, when their latitudes and longitudes are given.*

Let M and N be the two points whose latitudes and longitudes are given; P, the pole; and EQ, the equator. Join M to N by the arc of a great circle. Also let

E

M

R

P

through M and N arcs of great circles PMR and PNS, meeting the equator in R and S, we shall have

If we suppose PQ to be the meridian from which we reckon longitudes,

Then because PR = PS = 90°, we have in the triangle MPN

It is assumed in the solution of this problem that the earth is a sphere. Although the assumption is not strictly correct, the error to which it can give rise can never amount to more than a few thousandths of the whole distance.

This is a case, therefore, in which we have given two sides and the included angle to find the remaining side. If we put d for the required distance, the general formula gives

cos d = cos PM cos PN + sin PM sin PN cos P

= sin o sin 9' + cos o cos p' cos (λ — λ').

(a)

This equation will suffice to determine the distance between the points in arc of a great circle.

To reduce it to statute miles, the degrees must be multiplied by 691; and to reduce to nautical miles, by 60.

To solve the problem completely we should know not only the distance but the direction, or the angle which the great circle joining the two points makes with the meridian. This angle is different at the two points, being equal to the angle M of the spherical triangle at one point, and to N at the other. Hence the complete solution requires the complete solution of the spherical triangle PMN, for which we may use the Gaussian equations instead of the form (a).

, EXERCISES.

1. Find (1) the distance,* both in degrees and nautical miles, between New York and Liverpool, on an arc of a great circle; (2) the direction in which a ship would sail on leaving the one port for the other, on an arc of a great circle, and the direction in which she would be sailing on approaching her destination. The positions of the cities are:

New York........lat. + 40° 42'.7, long. 74° 0′.0 west. Liverpool......... lat. + 53° 24'.1, long. 2° 59′.1 west. 2. Compute the distance between Liverpool and Rio de Janeiro, the position of the latter being:

Note that the latitude of Rio is algebraically negative, being reckoned south from the equator.

3. If a ship sails from New York, starting due east, and continues her course on an arc of a great circle, what will be her latitude when she reaches the meridian of Greenwich, and in what direction will she then be sailing?

* By distance, as used here, distance on the arc of a great circle is to be understood, unless explicitly stated otherwise.

Geometrical Applications.

125. Def. Three straight lines, each perpendicular to the other two and all passing through a common point, are called a system of rectangular axes.

Def. The common point of intersection is called the origin. Def. Three planes, each perpendicular to the other two, are called three rectangular planes.

Remark. It is shown in geometry that three rectangular planes intersect each other in lines forming a system of rectangular axes.

If a sphere have its centre in the common point of intersection of three rectangular planes, these planes will intersect its surface in three great circles, forming in all eight trirectangular triangles.

126. THEOREM. The sum of the squares of the cosines of the three angles which a straight line forms with three rectangular axes is equal to unity.

This theorem is expressed in trigonometric language thus: If a, B, and y be the angles which a straight line forms with three rectangular axes, then

cos' acos B+ cos y = 1.

(1)

Proof. Let the line pass through the common point of intersection of the axes. Let a sphere have its centre at this point, and let X, Y, Z, and P be the points in which the axes and the line intersect the spherical surface.

Join PX, PY, and PZ by arcs of great circles, and produce ZP until it meets XY in Q. Then

X

Q

P

Because the angles a, 6, and y are formed at the centre of the

Because XYZ is a trirectangular triangle, Z is the pole of XY, and ZQ is therefore perpendicular to XY. Therefore the

spherical triangles XQP and YQP are right-angled at Q, and, by § 109, 5,

cos PX = cos a = cos XQ cos QP;

cos PY = cos B = cos YQ cos QP.

Taking the squares of these equations and adding them to the square of cos PZ = cos y, we have

cos'a + cos'ẞ+cos'y =cos' QP (cos' XQ + cos2 YQ) + cos2 PZ = cos2 QP+cos' PZ (because XY = 90°) = 1 (because QZ

=

90°). Q.E.D.

Another proof. Let OX, OY, and OZ be the axes meeting

in O, and let OP be the line

making with the lines OX, OY, OZ the respective angles

Z

R

P

Through P pass three planes

parallel to the respective planes

formed by the axes, taken two and two, namely,

Plane PSVR || plane XOY;

Plane PQTS || plane YOZ;

Plane PRWQ|| plane ZOX.

These planes will then form a rectangular parallelopiped OTQW-VSPR, of which OP will be a diagonal. By the property of this diagonal (Geom., § 692),

OP2 = OT2 + OW2 + O V3.

(a)

Moreover, because these planes are respectively parallel to the three rectangular planes, and because each of the axes is perpendicular to one of these planes, the three planes in question are each perpendicular to one of the axes.

If we join PT, PW, and PV, these lines, being in planes which, as just shown, are perpendicular to the axes OX, OY, and OZ, will be perpendicular to these axes (Geom.), and we shall have

Taking the sum of the squares of these three equations, and substituting for OT' + OW' + OV' its value (a), we have OP = OP (cos2 a + cos3ß + cos3 y).

Dividing by OP,

cos2 a + cos2 + cos y = 1. Q.E.D.

(2)

127. Corollary. It is easily shown that the angles a, ß, and y are the complements of the angles which OP forms with the three rectangular planes. For, pass a plane through P and OZ. Because OZ 1 plane XO Y, the cutting plane OZP is also perpendicular to XOY, and the line OQ in which it cuts XOY1 OZ. But, by definition of the inclination of a line to a plane (Geom., § 603), the angle POQ will be the inclination of OP to the plane XOY. Therefore, if we call this inclination a, we have

b and c being the inclinations of the line to the two other planes. Therefore sin a = cos a, sin b = cos ẞ, and sin c = cos y; whence

sin' a + sin b+sin' c = 1.

(3)

Because parallel lines have equal inclinations to a plane, the angles which any straight line makes with the three planes are equal to those made by a parallel to that line through 0. Hence:

The sum of the squares of the sines of the three angles which any straight line makes with three rectangular planes is equal to unity.

128. Case of a plane cutting three rectangular planes. We have the following theorem :

THEOREM. If any plane intersect three rectangular planes, the sum of the squares of the cosines of the three angles which it forms with them is equal to unity.

This can be demonstrated from the preceding theorem by dropping a perpendicular from the common point of intersection of the three rectangular planes upon the cutting plane. Because the rectangular axes and this line are each perpendicular to one of the four planes, the angles which they form with each other are