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8. If p, q, r be the arcs of great circles drawn from the angles of a triangle perpendicular to the opposite sides, (a, a), (B, B'), (y, v) the segments into which these arcs are divided, shew that

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9. In a spherical triangle if arcs be drawn from the angles to the middle points of the opposite sides, and if a, a' be the two parts of the one which bisects the side a, shew that

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10. The arc of a great circle bisecting the sides AB, AC of a spherical triangle cuts BC produced at Q; shew that

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11. If ABCD be a spherical quadrilateral, and the opposite sides AB, CD when produced meet at E, and AD, BC meet at F, the ratio of the sines of the arcs drawn from E at right angles to the diagonals of the quadrilateral is the same as the ratio of those from F.

12. If ABCD be a spherical quadrilateral whose sides AB, DC are produced to meet at P, and AD, BC at Q, and whose diagonals AC, BD intersect at R, then

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sin AB sin CD cos P ~ sin AD sin BC cos Q sin AC sin BD cos R.

13. If the arcs joining the extremities of the base of a spherical triangle with the middle points of the opposite sides are equal, the triangle is isosceles.

14. If the tangent of the radius of the circle described about a spherical triangle is equal to twice the tangent of the radius of the circle inscribed in the triangle, the triangle is equilateral.

15. The arc AP of a circle of the same radius as the sphere is equal to the greater of two sides of a spherical triangle, and the arc AQ taken in the same direction is equal to the less; the

sine PM of AP is divided at E, so that

EM

PM

=

the natural cosine

of the angle included by the two sides, and EZ is drawn parallel to the tangent to the circle at Q. Shew that the remaining side of the spherical triangle is equal to the arc QPZ.

16. If through any point P within a spherical triangle ABC great circles be drawn from the angular points A, B, C to meet the opposite sides at a, b, c respectively, prove that

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17. A and B are two places on the Earth's surface on the same side of the equator, A being further from the equator than B. If the bearing of A from B be more nearly due East than it is from any other place in the same latitude as B, what is the bearing of B from A?

18. From the result given in example 18 of Chapter v. infer the possibility of a regular dodecahedron.

19. A and B are fixed points on the surface of a sphere, and P is any point on the surface. If a and b are given constants, shew that a fixed point S can always be found, in AB or AB produced, such that

a cos AP+b cos BP = 8 cos SP,

where s is a constant.

20. A, B, C,... are fixed points on the surface of a sphere; a, b, c,... are given constants. If P be a point on the surface of the sphere, such that

a cos AP + b cos BP + c cos CP + constant, shew that the locus of P is a circle.

XIV. NUMERICAL SOLUTION OF SPHERICAL TRIANGLES.

166. We shall give in this Chapter examples of the numerical solution of Spherical Triangles.

We shall first take right-angled triangles, and then obliqueangled triangles.

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168. Given A = 55° 32′ 45′′, C = 90°, c = 98° 14′ 24′′.

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Here cos c is negative; and therefore cot B will be negative, and B greater than a right angle. The numerical value of cos c is the same as that of cos 81° 45′ 36′′.

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Here tan c is negative; and therefore tan b will be negative and b greater than a quadrant,

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169. Given A = 46° 15′ 25′′, C = 90°, a = 42° 18′ 45′′,

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