Page:Spherical Trigonometry (1914).djvu/24

6 11. If the arcs of great circles joining a point $$\mathrm{P}$$ on the surface of a sphere with two other points $$\mathrm{A}$$ and $$\mathrm{C}$$ on the surface of the sphere, which are not at opposite extremities of a diameter, be each of them equal to a quadrant, $$\mathrm{P}$$ is a pole of the great circle through $$\mathrm{A}$$ and $$\mathrm{C}$$. (See the figure of Art. 7.)

For suppose $$PA$$ and $$PC$$ to be quadrants, and $$O$$ the centre of the sphere; then since $$PA$$ and $$PC$$ are quadrants, the angles $$POC$$ and $$POA$$ are right angles. Hence $$PO$$ is at right angles to the plane $$AOC$$, and $$P$$ is a pole of the great circle $$AC$$.

12. Definition. Great circles which pass through the poles of a great circle are called secondaries to that circle. Thus, in the figure of Art. 8 the point $$C$$ is a pole of $$ABMN$$, and therefore $$CM$$ and $$CN$$ are parts of secondaries to $$ABMN$$. And the angle between $$CM$$ and $$CN$$ is measured by $$MN$$; that is, the angle between any two great circles is measured by the arc they intercept on the great circle to which they are secondaries.

13. If from a point on the surface of a sphere there can be drawn two arcs of great circles, not parts of the same great circle, the planes of which are at right angles to the plane of a given circle, that point is a pole of the given circle.

For, since the planes of these arcs are at right angles to the plane of the given circle, the line in which they intersect is perpendicular to the plane of the given circle, and is therefore the axis of the given circle; hence the point from which the arcs are drawn is a pole of the circle.

14. To compare the arc of a small circle subtending any angle at the centre of the circle with the arc of a great circle subtending the same angle at its centre.

Let $$ab$$ be the arc of a small circle, $$C$$ the centre of the circle, $$P$$ the pole of the circle, $$O$$ the centre of the sphere. Through $$P$$ draw the great circles $$PaA$$ and $$PbB$$, meeting the great circle of which $$P$$ is a pole, at $$A$$ and $$B$$ respectively; draw $$Ca$$, $$Cb$$, $$OA$$,