It may be shewn by common trigonometrical formulæ, that

b)

4 sina sin b sin &c=sin (sa) + sin (8-6) + sin (sc) - sin 8;

hence we have from (4)

17

2n

{sin ( s − a) + sin (s — b) + sin (s — c) — si s ........5).

93. To find the angular radii of the small circles described round the triangles associated with a given fundamental triangle.

1

2

3

Let R, denote the radius of the circle described round the triangle formed by producing AB and AC to meet again in A'; similarly let R, and R, denote the radii of the circles described round the other two triangles which are similarly formed. Then we may deduce expressions for tan R1, tan R2, and tan R those found in Art. 92 for tan R. The sides of the triangle ABC are а, π-b, π· c, and its angles are A, π-B, π-C; hence if § (a + b + c) and S = 1⁄2 (A + B + C) we shall obtain from

8 =

Art. 92

=

tana

1 cos S

tan R1

1

2n

sin 8 — sin (8 − a) + sin (8 — 6) + sin ( )...........

Similarly we may find expressions for tan R, and tan R.

2

94. Many examples may be proposed involving properties of the circles inscribed in and described about the associated triangles. We will give one that will be of use hereafter.

4n2 = 1 - cos2 a - cos2 b- cos c + 2 cos a cos b cos c;

-

c).

sin 8 + sin (8 − a) + sin (8 − 6) + sin (s — c)};

members of this equation the required For it may be shewn by reduction that

sin2 s + sin2 (s − a) + sin3 (s − b) + sin3 (s—c) = 2 − 2 cos a cos b cos c,

and

sins sin (sa) + sin s sin (s—b) + sin s sin (s—c)

+ sin (s - a) sin (s — b) + sin (s — b) sin (s − c) + sin (s — c) sin (s − a)

= sin a sin b + sin b sin c + sin c sin a.

95. In the figure to Art. 89, suppose DP produced through P to a point A' such that DA' is a quadrant, then A' is a pole of

π

BC and PA'=-r; similarly, suppose EP produced through P

2

to a point B such that EB' is a quadrant, and FP produced through P to a point C' such that FC' is a quadrant.

A'B'C' is the polar triangle of ABC, and PA' = PB′ = PC′ :

Thus P is the pole of the small circle described round the polar triangle, and the angular radius of the small circle described round the polar triangle is the complement of the angular radius of the

T. S. T.

F

small circle inscribed in the primitive triangle. And in like manner the point which is the pole of the small circle inscribed in the polar triangle is also the pole of the small circle described round the primitive triangle, and the angular radii of the two circles are complementary.

EXAMPLES.

In the following examples the notation of the chapter is retained.

Shew that in any triangle the following relations hold contained in Examples 1 to 5:

6. Shew that in an equilateral triangle tan R = 2 tan r.

7. If ABC be an equilateral spherical triangle, P the pole of the circle circumscribing it, Q any point on the sphere, shew that

cos QA + cos QB + cos QC = 3 cos PA cos PQ.

8. If three small circles be inscribed in a spherical triangle having each of its angles 120°, so that each touches the other two as well as two sides of the triangle, prove that the radius of each of the small circles 30°, and that the centres of the three small circles coincide with the angular points of the polar triangle.

=

VIII. AREA OF A SPHERICAL TRIANGLE.

SPHERICAL EXCESS.

96. To find the area of a Lune.

A Lune is that portion of the surface of a sphere which is com prised between two great semicircles.

CDEE

B

Let ACBDA, ADBEA be two lunes having equal angles at A; then one of these lunes may be supposed placed on the other so as to coincide exactly with it; thus lunes.having equal angles are equal. Then by a process similar to that used in the Sixth Book of Euclid it may be shewn that lunes are proportional to their angles. Hence since the whole surface of a sphere may be considered as a lune with an angle equal to four right angles, we have for a lune with an angle of which the circular measure is A,

Suppose

the radius of the sphere, then the surface is 4

(Integral Calculus, Chap. VII.);

thus

97. To find the area of a Spherical Triangle.

The

Let ABC be a spherical triangle; produce the arcs which form its sides until they meet again two and two, which will happen when each has become equal to the semi-circumference. triangle ABC now forms a part of three lunes, namely, ABDCA, BCEAB, and CAFBC. Now the triangles CDE and AFB are subtended by vertically opposite solid angles at O, and we will assume that their areas are equal; therefore the lune CAFBC is equal to the sum of the two triangles ABC and CDE. Hence if A, B, C denote the circular measures of the angles of the triangle, we have

triangle ABC + BGDC = lune ABDCA = 2Ar2,

triangle ABC + AHEC = lune BCEAB = 2Br2,

triangle ABC + triangle CDE = lune CAFBC = 2Cr2 ;

hence, by addition,

twice triangle ABC + surface of hemisphere = 2 (A + B + C) r2 ; therefore triangle ABC = (A + B + C − π) r2.

The expression A+B+C-π is called the spherical excess of