and her departure 2425 miles. Required her course, distance and longitude.

Ans. Course N. 79° 37′ W.

Distance 2465.3 miles.

Longitude 67° 9'.3 W.

7. Find the bearing and distance from Sandy Hook, latitude 40° 28' N., longitude 74° 1' W., to the Cape of Good Hope, latitude 34° 22' S., longitude 18° 30' E.

Ans. Course

Distance

CHARTS.

(204.) The charts commonly used in navigation are plane charts, or Mercator's chart. In the construction of the former, the portion of the earth's surface which is represented is supposed to be a plane. The meridians are drawn parallel to each other, and the lines of latitude at equal distances. The distance between the parallels should be to the distance between the meridians, as radius to the cosine of the middle latitude of the chart. A chart of moderate extent constructed in this manner will be tolerably correct. The distance of the meridians in the middle of the chart will be exact, but on each side it will be either too great or too small.

When large portions of the earth's surface are to be represented, the error of the plane chart becomes excessive. To obviate this inconvenience Mercator's chart has been constructed. Upon this chart the meridians are represented by parallel lines, and the distance between the parallels of latitude is proportioned to the meridional difference of latitude, as represented on page 149.

We have seen that the meridional difference of latitude is to the difference of longitude as radius is to the tangent of the course. Hence, while the course remains unchanged, the ratio of the meridional difference of latitude to the difference of longitude is constant; and, therefore, every rhumb line will be represented on Mercator's chart by a straight line. This property renders Mercator's chart peculiarly convenient to navigators

The preceding sketch affords a very incomplete view of the present state of the science of navigation. The most accurate method of ascertaining the situation of a vessel at sea is by means of astronomical observations. For these, however, the student must be referred to some treatise on Astronomy.

BOOK VI.

SPHERICAL TRIGONOMETRY.

(205.) SPHERICAL trigonometry teaches how to determine the several parts of a spherical triangle from having certain parts given.

A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles, each of which is less than a semicircumference.

RIGHT-ANGLED SPHERICAL TRIANGLES.

THEOREM I.

(206.) In any right-angled spherical triangle, the sine of the hypothenuse is to radius, as the sine of either side is to the sine of the opposite angle.

Let ABC be a spherical triangle, right-angled at A; then will the sine of the hypothenuse BC

be to radius, as the sine of the side

straight line EF, in the plane ABD, perpendicular to BD, and join CF. Then, because DB is perpendicular to the two lines CE, EF, it is perpendicular to the plane CEF; and, consequently, the plane CEF is perpendicular to the plane ABD (Geom., Prop. 6, B. VII.). But the plane CAD is also perpendicular to the plane ABD; therefore their line of common section, CF, is perpendicular to the plane ABD; hence CFD, CFE are right angles, and CF is the sine of the arc AC. Now, in the right-angled plane triangle CFE,

CE: radius:: CF: sine CEF.

But since CE and FE are both at right angles to DB, the angle CEF is equal to the inclination of the planes CBD, ABD; that is, to the spherical angle ABC. Therefore,

sine BC R:: sine AC: sine ABC.

(207.) Cor. 1. In any right-angled spherical triangle, the sines of the sides are as the sines of the opposite angles. For, by the preceding theorem,

and

sine BC: R: sine AC: sine ABC,

sine BC: R: sine AB : sine ACB; therefore, sine AC: sine AB :: sine ABC: sine ACB.

Cor. 2. In any right-angled spherical triangle, the cosine of either of the sides is to radius, as the cosine of the hypothenuse is to the cosine of the other side.

F

E

Let ABC be a spherical triangle, right-angled at A. Describe the circle DE, of which B is the pole, and let it meet the three sides of the triangle ABC produced in D, E, and F. Then, because BD and BE are quadrants, the arc DF is perpendicular to BD. And since BAC is a right angle, the arc AF is perpendicular to BD. Hence the point F, where the arcs FD, FA intersect each other, is the pole of the arc BD (Geom., Prop. 5, Cor. 2, B. IX.), and the arcs FA, FD are quadrants.

B

D

A

Now, in the triangle CEF, right-angled at the point E, according to the preceding theorem, we have

sine CF R :: sine CE : sine CFE.

But CF is the complement of AC, CE is the complement of BC, and the angle CFE is measured by the arc AD, which is the complement of AB. Therefore, in the triangle ABC, we have

cos. AC: R:: cos. BC: cos. AB.

Cor. 3. In any right-angled spherical triangle, the cosine of either of the sides is to radius, as the cosine of the angle opposite to that side is to the sine of the other angle.

For, in the triangle CEF, we have

sine CF: R: sine EF sine ECF.

But sine CF is equal to cos. CA. EF is the complement of

ED, which measures the angle ABC, that is, sine EF is equal to cos. ABC, and sine ECF is the same as sine ACB; therefore,

cos. AC: R:: cos. ABC : sine ACB

THEOREM II.

(208.) In any right-angled spherical triangle, the sine of either of the sides about the right angle is to the cotangent of the adjacent angle, as the tangent of the remaining side is to radius.

Let ABC be a spherical triangle, right-angled at A; then will the sine of the side AB be to the cotangent of the angle ABC, as the tangent of the side AC is to radius.

Let D be the center of the sphere; join AD, BD, CD; draw AE perpendicular to BD, which will, therefore, be the sine of the arc AB. Also, from Dthe point E in the plane BDC, draw

the straight line EF perpendicular to

BD, meeting DC produced in F, and

E

B

A

join AF. Then will AF be perpendicular to the plane ABD, because, as was shown in the preceding theorem, it is the com mon section of the two planes ADF, AEF, each perpendicular to the plane ADB. Therefore FAD, FAE are right angles, and AF is the tangent of the arc AC.

Now, in the triangle AEF, right-angled at A, we have
AE: radius::AF: tang. AEF.

But AE is the sine of the arc AB, AF is the tangent of tkarc AC, and the angle AEF is equal to the inclination of tha planes CBD, ABD, or to the spherical angle ABC; hence sine AB R:: tang. AC: tang. ABC.

And because, Art. 28,

R: cot. ABC: tang. ABC: R; therefore, sine AB: cot. ABC :: tang. AC: R.

(209.) Cor. 1. In any right-angled spherical triangle, she cosine of the hypothenuse is to the cotangent of either of the oblique angles, as the cotangent of the other oblique ange is to radius.