By means of the polar triangle, we may deduce from the result just found, that

(4) Required the relation between the small variations of a side and the adjacent angle (b, A).

We have

cot C sin A+ cos b cos A.

(Art. 193)

cot c sin b = Giving to b and A very small increments, and subtracting, as before, we get

cot c cos bdb = cot C cos AdA sin bcos Adb ·

cos b sin AdA.

(cotc cos b + sin b cos A) db = (cot C cos A - cos b sin A)dA.

1. If A and c are constant, prove the following relations between the small variations of any two parts of the other

2. If B and C remain constant, prove the following:

POLYEDRONS.

231. To find the Inclination of Two Adjacent Faces of a Regular Polyedron.

Let C and D be the centres of the circles inscribed in the two adjacent faces whose common edge is AB; bisect AB in E, and join CE and DE; CE and DE C will be perpendicular to AB. .. ≤ CED is the inclination of the two adjacent faces, which denote by I.

B

In the plane CED draw CO and DO at right angles to CE and DE, respectively, and meeting in O. Join OA, OE, OB, and from O as centre describe a sphere, cutting OA, OC, OE at a, c, e, respectively; then ace is a spherical triangle. Since AB is perpendicular to CE and DE, it is perpendicular to the plane CED; therefore the plane AOB, in which AB lies, is perpendicular to the plane CED. .. aec is a right angle.

Let m be the number of sides in each face, and n the number of plane angles in each solid angle. Then

Cor. 1. If r be the radius of the inscribed sphere, and a be a side of one of the faces, then

For,

r=OC=CE tan CEO=AE cot ACE tan CEO

Cor. 2. If R be the circumradius of the polyedron, then

For, r=OA cos aoc = R cot eca cot eac = R cot cotπ.

π

Cor. 3. The surface of a regular polyedron, F being the

For, the volume of the pyramid which has one face of the polyedron for base and O for vertex

232. Volume of a Parallelopiped. — To find the volume of a parallelopiped in terms of its edges and their inclinations to one another.

piped is equal to the area of the base OAEB multiplied by the altitude CH; that is,

volume =ab sin y⚫ CH

= abc sin y sin

ce

where ce is the perpendicular arc from c on ab.

... volume = abc sin y sin ac sin bac

(Art. 186)

(Art. 195)

sin ẞ sin y

=

abc √1-cos2 α

cos2 B-cos2y + 2 cos a cos ẞ cos y.

Cor. 1. The surface of a parallelopiped

= 2 (bc sin a + ca sin ß + ab sin y).

Cor. 2. The volume of a tetraedron

=abc √1-cos2 a-cos2 ß-cos2y+2 cos a cos ẞcos y.

For, a tetraedron is one-sixth of a parallelopiped which has the same altitude and its base double that of the tetraedron.

233. Diagonal of a Parallelopiped. To find the diagonal of a parallelopiped in terms of its edges, and their mutual inclinations.

Let OD (figure of Art. 232) be a parallelopiped, whose edges OA = a, OB = b, OC = c, and their inclinations BOC =α, COA = B, AOB = y; let OD be the diagonal required,

and OE the diagonal of the face OAB.

OED gives

Then the triangle

OD2 = OE2 + ED2 + 2 OE. ED cos COE

= a2 + b2 + 2 ab cos y + c2 + 2c · OE cos COE (1) Now, it is clear that OE cos COE is the projection of OE on the line OC, and therefore it must be equal to the sum of the projections of OB and BE (or of OB and OA), on the same line.*

.. OE cos COE =

which in (1) gives

b cos a + a cos ß,

OD2 = a2 + b2 + c2 + 2 bc cos a + 2 ca cos B + 2 ab cos y. (2)

234. Table of Formulæ in Spherical Trigonometry. For the convenience of the student, many of the preceding formulæ are summed up in the following table:

10. cos c = cos a cos b + sin a sin b cos C.

* From the nature of projections (Plane and Solid Geom., Art. 326).