Elements of Geometry and Trigonometry

Such are the very simple expressions, by means of which the inclination of two faces in all the five regular polyedrons is determined. We shall farther observe, that they might all have been included under one single formula.

Thus, let n be the number of sides in each face, m the number of plane angles combined in each solid angle; if from the point O with a radius =1, a spherical surface be described to meet the line OA, OC, OD in p, q, r, we shall have a spherical triangle pqr, in which are known the

angle r, the angle p ==, and the angle q=;

hence by admitted formulas, we shall have cos P. But cos qrcos COD = sin CDO

cos qr=

sin q

sin C, C designating the angle CDE; hence

COS

sin C; a general formula which, if applied successively to the

sin

T
n

five polyedrons, will give the same values of cos C, or of 1-sin2 C, as were found above by another method. For this purpose, we must substitute the values of m and n in each particular case, namely,

The same spherical triangle pqr, from which the inclination of two adjacent faces has now been deduced, gives also cos pq=cot p cot q,,

со

cot. Hence, if the radius of the sphere circum

scribed about the polyedron is called R, the radius of the sphere inscribed in the same polyedron being called r, we shall have

Ja tang; also, making AB=a, we have CA=

polyedron, the values of R and r, the radii of the circumscribed and of the inscribed sphere. Likewise, supposing C known, we have r= a cottang C, and Ra tang tang ¿C.

same value, tangtang Hence if R is the same for both, r also

will be the same; in other words, if these two solids are inscribed in the same sphere, they will also be circumscribed about the same sphere, and vice versa. The same property subsists between the hexaedron

Ꭱ

and the octaedron, the value of being tang tang, for both.

We may observe, that the regular polyedrons are not the only solids contained by equal and regular polygons. If two regular tetraedrons are adjusted to each other by a common face, a solid bounded by six equal and equilateral triangles will result. Another solid might be formed with ten equal and equilateral triangles; but the regular polyedrons are the only ones which at the same time have their solid angles equal.

NOTE X.

On the Area of the Spherical Triangle.

LET 1 be the radius of the sphere, the semicircumference of a great circle; let a, b, c be the three sides of a spherical triangle; A, B, C the arcs of great circles which measure the opposite angles. Put A+B+C S. According to what was proved in the text (Prop. 23. VII.), the area of the spherical triangle is equal to the arc S multiplied by the radius, and consequently it is represented by S. Now, by Napier's analogies, we have,

from which having found the value of tang (A+B), we shall easily deduce that of tang (↓ A+B+C)=—cot S: we shall thus have cot la cot b+cos C sin C

cot &S=

a very simple formula, which will serve for calculating the area of a spherical triangle, when two sides a, b, and the included angle C, are known. Some remarkable consequences may also be deduced from it.

First. If the angle C is constant, and likewise the product cot

cot the area of the spherical triangle represented by S will also be 2'

constant.

Hence two triangles CAB, CDE, which have an equal angle

C, will be equivalent, if they have

CB,

tang CA tang CD:: tang CE: tang
in other words, if the tangents of the half-sides
containing the equal angles are reciprocally pro-
portional.

Second. On the given side CD, and with the same angle C, to make a triangle CDE equivalent to the given triangle CAB, we must determine CE by the proportion

tang CE tang CA :: tang CB: tang CE. Third. With the angle C at the vertex, to make an isosceles triangle DCE equivalent to the given triangle CAB, we must take tang CD, or tang CE, a mean proportional between tang CA and tang CB. cot a cot b+cos C

Fourth. The same formula cot S

sin C

will af

ford a very simple demonstration of Prop. 20. Book VII., where it is proved that of all the spherical triangles formed with two given sides a and b, the greatest is the one in which the angle C, included between the given sides, is equal to the sum of the other two angles A and B.

With the radius OZ = 1, describe the semicircumference VMZ; make the arc ZX-C; and on the other side of the centre, take OP = cot a cot b; lastly join PX, and draw XY perpendicular to PZ.

Y Z

In the right-angled triangle PXY, we have cot P =

cot la cot b + cos C sin C

; hence PS; hence the surface S will be a maximum, when the angle P is one. Now it is evident that if PM is drawn to touch the circumference, the angle MPO will be the maximum of the angles P; in which case, we shall have MPO MOZ — }; Hence the spherical triangle with two given sides, will be a maximum, if we have SC — 1, or C =A+B; which agrees with the Proposition referred to.

From this construction, it is also plain that no maximum could exist, if the point P were within the circle, that is, if we had cot la cot b1; a condition from which we successively deduce cotatang 3 b, tang (a) — tang 1b, 1π — a 16, and finally * <a+b; which also agrees with the Scholium of the same Proposition.

PROBLEM I. To find the surface of a spherical triangle by means of

its three sides.

To effect this, in the formula

cot la cot b + cos C

cot &S=

sin C

we must substitute the values of sin C and cos C expressed in terms of