SOLID ANGLES.

BOOK VI.

DEFINITIONS.

1. A solid angle is the angular space contained between several planes which meet in the same point.

2. Three planes, at least, are required to form a solid angle. 3. A solid angle is called a trihedral, tetrahedral, &c., an gle, according as it is formed by three, four,.... plane angles.

4. If three planes intersect each other in a common point, the indefinite space bounded by those three planes is called a triangular pyramidal space.

5. If three planes intersect each other in parallel lines, the indefinite space bounded by those three planes is called a triangular prismatic space.

6. If four or more planes intersect each other in a common point, the indefinite space bounded by those planes is called a polyhedral pyramidal space.

7. If four or more planes intersect each other in parallel edges, the indefinite space bounded by them is called a polyhedral prismatic space.

PROPOSITION I.

If a solid angle be contained by three plane angles, the sum of any two of these angles will be greater than the third.

It is unnecessary to demonstrate this proposition except in the case where the plane angle which is compared with the two others is greater than either of them..

Let A be a solid angle contained by the three plane angles BAC, CAD, DAB, and let BAC be the greatest of these an

gles;

C

E

D

Then,

CAD+DAB>BAC.

In the plane BAC draw the straight line AE, making the angle BAE angle BAD.

Make AE AD, and through E draw any straight line BEC, cutting AB, AC, in the points B, C; join D, B; D, C; Then, AD AE, and AB is common to the two triangles DAB, BAE, and the angle DAB-angle BAE,

But

.. BD-BE. BD+DC>BE+EC, .. DC>EC.

Again, AD=AE, and AC is common to the two triangles DAC, EAC, but the base DC>base EC,

But

.. angle DAC>angle EAC.

angle DAB-angle BAE.

:. angle CAD+angle DAB>angle BAE+angle EAC

>angle BAC.

PROPOSITION II.

The sum of the plane angles which form a solid angle is

always less than four right angles.

Let P be a solid angle contained by any number of plane angles APB, BPC, CPD, DPE, EPA.

Let the solid angle P be cut by any plane ABCDE.

Take any point, O, in this plane; join A, O; B, O; C, O; D, O; E, O;

Then, since the sum of all the angles of every triangle is always equal to two right

P

E

A

B

angles, the sum of all the angles of the triangles ABP, BBC,

... about the point P, will be equal to the sum of all the angles of the equal number of triangles AOB, BOC,...... about the point O.

Again, by the last proposition, angle ABC<angle ABP +angle CBP; in like manner, angle BCD<angle BCP +angle DCP, and so for all the angles of the polygon ABCDE.

Hence the sum of the angles at the bases of the triangles whose vertex is O is less than the sum of the angles at the bases of the triangles whose vertex is P.

.. The sum of the angles about the point O must be greater than the sum of the angles about the point P. But the sum of the angles about the point O is four right angles.

The sum of the angles about the point P is less than four right angles.

PROPOSITION III.

If two solid angles be formed by three plane angles which are equal, each to each, the planes in which these angles lie will be equally inclined to each other.

Let P, Q, be two solid angles,

Р

each contained by three plane angles;

Let angle APC-angle DQF,

angle APB angle DQE, and an- A/B

gle BPC angle EQF.

Then, the inclination of the

D/E

Z

F

planes APC, APB, will be equal to the inclination of the planes DQF, DQE.

Take any point, B, in the intersection of the planes APB, CPB.

From B draw BY perpendicular to the plane APC, meeting the plane in Y.

From Y draw YA, YC, perpendiculars on PA, PC; join A, B; B, C;

Again, take QE=PB, from E draw EZ perpendicular to the plane DQF, meeting the plane in Z; from Z draw ZD, ZF, perpendiculars on QD, QF; join D, E; E, F.

The triangle PAB is right angled at A, and the triangle QDE is right angled at D. (Geom. of Planes, Prop. 7.) Also, the angle APB-angle DQE, by construction.

.. angle PBA=angle QED.

But the side PB-side QE,.. the two triangles APB, DQF, are equal and similar.

.. PA=QD, and AB=DE.

In like manner, we can prove that

PC=QF, and BC=EF.

We can now prove that the quadrilateral PAYC is equal to the quadrilateral QDZF.

For, let the angle APC be placed upon the equal angle DQF, then the point A will fall upon the point D, and the point C on the point F, because PA-QD, and PC=QF.

At the same time, AY, which is perpendicular to PA, will fall upon DZ, which is perpendicular to QD; and, in like manner, CY will fall upon FZ.

Hence the point Y will fall on the point Z, and we shall have

But the triangles AYB, DZE, are right angled at Y and Z, the hypothenuse AB-hypothenuse DE, and the side AY -side DZ; hence these two triangles are equal.

.. angle YAB-angle ZDE.

The angle YAB is the inclination of the planes APC, APB; and

The angle ZDE is the inclination of the planes DQF, DQE. ..these planes are equally inclined to each other.

In the same manner, we prove that angle YCB=angle ZFE, and, consequently, the inclination of the planes APC, BPC, is equal to the inclination of the planes DQF, EQF. .

We must, however, observe that the angle A of the rightangled triangle YAB is not, properly speaking, the inclination of the two planes APC, APB, except when the perpendicular BY falls upon the same side of PA as PC does; if it fall upon the other side, then the angle between the two planes will be obtuse, and, added to the angle A of the triangle YAB, will make up two right angles. But, in this case, the angle between the two planes DQF, DQE, will also be obtuse, and, added to the angle D of the triangle ZDE, will make up two right angles.

Since, then, the angle A will always be equal to the angle D, we infer that the inclination of the two planes APC, APB, will always be equal to the inclination of the two planes DQF, DQE. In the first case, the inclination of the plane is the angle A or D; in the second case, it is the supplement of those angles.

SCHOLIUM.-If two solid trihedral angles have the three plane angles of the one equal to the three plane angles of the other, each to each, and, at the same time, the corresponding angles arranged in the same manner in the two sol

id angles, then these two solid angles will be equal; and if placed one upon the other, they will coincide. In fact, we have already seen that the quadrilateral PAYC will coincide with the quadrilateral QDZF. Thus the point Y falls upon the point Z, and, in consequence of the equality of the triangles AYB, DZE, the straight line YB, perpendicular to the plane APC, is equal to the straight line ZE, perpendicular to the plane DQE; moreover, these perpendiculars lie in the same direction; hence the point B will fall upon the point E, the straight line PB on the straight line QE, and the two solid angles will entirely coincide with each other. This coincidence, however, can not take place except we suppose the equal plane angles to be arranged in the same manner in the two solid angles; for, if the equal planes be arranged in an inverse order, or, which comes to the same thing, if the perpendiculars YB, ZE, instead of being situated both on the same side of the planes APC, DQF, were situated on opposite sides of these planes, then it would be impossible to make the two solid angles coincide with each. other. It would not, however, be less true, according to the above theorem, that the planes in which the equal angles lie would be equally inclined to each other; so that the two solid angles would be equal in all their constituent parts, without admitting of superposition. This species of equality, which is not absolute, or equality of coincidence, has received from Legendre a particular description. He terms it equality of symmetry.

Thus the two solid trihedral angles in question, which have the three plane angles of the one equal to the three plane angles of the other, each to each, but arranged in an inverse order, are termed angles equal by symmetry, or simply symmetrical angles.

The same observation applies to solid angles formed by more than three plane angles. Thus a solid angle formed by the plane angles A, B, C, D, E, and another solid angle formed by the same angles in an inverse order, A, E, D, C, B, may be such that the planes in which the equal angles are situated are equally inclined to each other. These two solid angles, which would in this case be equal, although not admitting of superposition, would be termed solid angles equal by symmetry, or symmetrical solid angles.