Similar formulæ of course hold for the points of intersection of the touching circle with the other sides. 145. Let x denote the perpendicular from the pole of the touching circle on AB; then A sin z= sin B sin (* + ) = sin B (sin + oon ain y). A А 2 Therefore COS sin %= A A Z a n 008 ° + +2/cos4–sin sin, sin } Now Zain" sin (8 – a) sin brind) sin (8-4){ (9 - 0 1 - eoe a com von ਨੂੰ · °12 A ) (c c 6 COS COS S sec is equal to the product of sin (8 - a) sin (8-8) sin b sinc into 6 sin (8–c) + sins sin (8-c) cos 8 + cos (8 –c) sin 2 2 a sec COS COS {s 20} sin (8 – a) sin (3 – 5) {2si 2-in-..., con o sin (26-2) 28 –c} sin (–a) sin (8 –1}{2sinat 2 sin b sin = sin p cos (A - B). .P Similar expressions hold for the perpendiculars from the pole of the touching circle on the other sides of the spherical triangle. 146. Let P denote the point determined in Art. 139; G the point determined in Art. 140, and N the pole of the touching circle. We shall now shew that P, G, and N are on a great circle. Let x, y, z denote the perpendiculars from N on the sides a, b, c respectively of the spherical triangle ; let X, Y, Z, denote the perpendiculars from P; and x, y, w, the perpendiculars from Y 1' where t, and t, are certain quantities the values of which are not required for our purpose. Therefore by Art. 137 a certain point in the same great circle as P and G is at the perpendicular distances x, y, z from the sides a, b, c respectively of the spherical triangle: and hence this point must be the point N. 147. The resemblance of the results which have been obtained to those which are known respecting the Nine points circle in Plane Geometry will be easily seen, 1 The result tan p= = R tan R corresponds to the fact that the radius of the Nine points circle is half the radius of the circumscribing circle of the triangle. From equation (4) of Art. 144 by supposing the radius of the 78 + C - a sphere to become infinite we obtain : this corresponds to the fact that the Nine points circle passes through the feet of the perpendiculars from the angles of a triangle on the opposite sides. From equation (5) of Art. 144 by supposing the radius of the sphere to become infinite we obtain = this corresponds to the 2 : fact that the Nine points circle passes through the middle points of the sides of a triangle. From Art, 145 by supposing the radius of the sphere to be 1 come infinite we obtain 2= Rcos (A – B): this is a known a 2 property of the Nine points circle. In Plane Geometry the points which correspond to the P, G, and N of Art. 146 are on a straight line. 2c C M 148. The results which have been demonstrated with respect to the circle which touches the inscribed and escribed circles of a spherical triangle are mainly due to Dr Hart and Dr Salmon. See the Quarterly Journal of Mathematics, Vol. vr. page 67, EXAMPLES a VE cos Scos (S – A) 2 1. From the formula sin deduce 2 sin B sin C the expression for the area of a plane triangle, namely aʻ sin B sin c when the radius of the sphere is indefinitely in2 sin A creased. 2. Two triangles ABC, abc, spherical or plane, equal in all respects, differ slightly in position : shew that cos A Bb cos BCc cos CAa + cos ACc cos C Bb cos BAa=0. 3. Deduce formulæ in Plane Trigonometry from Napier's Analogies. 4. Deduce formulæ in Plane Trigonometry from Delambre's Analogies. a +6 COS C A + B С 5. From the formula cos COS sin deduce 2 2 2 2 the area of a plane triangle in terms of the sides and one of the angles. 6. What result is obtained from Example 7 to Chapter VI., by supposing the radius of the sphere infinite ? From the angle C of a spherical triangle a perpendicular is drawn to the arc which joins the middle points of the sides a and b: shew that this perpendicular makes an angle S-B with the side a, and an angle S- A with the side b. 8. From each angle of a spherical triangle a perpendicular is drawn to the arc which joins the middle points of the adjacent sides. Shew that these perpendiculars meet at a point; and that |