angle, which may be called a right solid angle, will serve as a very natural unit of measure for all other solid angles; and, if so, the same number that exhibits the area of a spherical polygon, will exhibit the measure of the corresponding solid angle. If the area of the polygon is 3, for example, in other words, if the polygon is of the trirectangular polygon, then the corresponding solid angle will also be of the right solid angle.

PROPOSITION XXXVI.

THEOREM. The surface of a spherical polygon is measured by the sum of all its angles, minus the product of two right angles by the number of sides in the polygon minus

two.

From one of the vertices A, let diagonals AC, AD be drawn to all the other vertices; the polygon ABCDE will be divided into as many E triangles, minus two, as it has sides. But the surface of each triangle is measured by the sum of all its

D

A

B

angles minus two right angles, and the sum of the angles in all the triangles is evidently the same as that of all the angles in the polygon: hence the surface of the polygon is equal to the sum of all its angles, diminished by twice as many right angles as it has sides minus two.

Schol. Let s be the sum of all the angles in a spherical polygon, and n the number of its sides; the right angle being taken for unity, the surface of the polygon will be measured by s-2 (n-2), or s-2n+4.

APPENDIX.

APPLICATION OF ALGEBRA TO THE SOLUTION OF GEOMETRICAL ~

PROBLEMS.

PROBLEM I. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a certain point within it to the three sides, to determine its side.

C

F

E

Let ABC be the equilateral triangle, and DE, DF, DG the perpendiculars from the point D upon the sides respectively. Denote these perpendiculars by a, b, c, in order, and the side of the triangle ABC by 2x. Then, if the perpendicular CH be drawn, CH √AC'—AH' = √4x'—x2 = x√3.

A

B

H G

The area of the triangle ADB = AB.GD = cx. Similarly the triangle BDC= ax, the triangle CDA = bx, and the triangle ACB = AB. CH = x2√3. Also BDC + CDA + ADB ABC; that is, in symbols,

=

we also had CH

x√3 = a+b+c:

= x√3. Hence CH a + b + c ; or

the whole perpendicular CH is equal to the sum of the three smaller perpendiculars from D upon the sides, whenever the point D is taken within the triangle. Had the point D been taken without the triangle, the perpendicular upon the side which subtends the angle within which the point lies would become negative. Thus, had the point been without the triangle, but between the sides AB, AC produced, then CH = DF + DG DE.

PROBLEM II. A maypole was broken off by the wind, and its top struck the ground twenty feet from the base; and, being repaired, was broken a second time five feet lower, and its top struck the ground ten feet farther from the base. What was the height of the maypole? Let AB be the unbroken maypole, C

and H the points in which it was successively broken, and D and F the corresponding points at which the top B struck the ground. Then will CAD and HAF be right-angled triangles.

Put BC CD=x, CA=y, AD-a, AF6, and CH c. Then AB=x+y, BH HF=x+c, and HA= =y-c; therefore [B. II, Prop. vII] we have

ין

1

IB

CH

Expanding [2] and subtracting [1] from it, we have, after a slight reduction,

b2-a2

x+y =

50 feet, the required height.

20

PROBLEM III. A statue eighty feet high stands on a pedestal fifty feet high, and, to a spectator on the horizontal plane, they subtend equal angles; required the distance of the observer from the base, the height of the eye being five feet.

Let AB =a, the height of the

sought. Then EC'

the distance

=

B

EF'+CF2 = x2+(a+b-c)', and

EA EF2+ED2 = x2+c'.

But, since the angle CEB = BEA, we have [B. IV, Prop. EC EA :: CB: BA, or

XIII]

EC2 EA :: CB2: BA'; or, in symbols,

x2+(a+b-c) : x2+c2 :: b2: a'.

From this proportion, we readily deduce

the double sign merely indicating that this value of x may be measured either way, from A towards D, or from D towards A.

PROBLEM IV. Given the three sides a, b, c of a triangle, to find:

1. The three perpendiculars from the angles upon the opposite sides;

2. The area of the triangle;

3. The radius of the circumscribed circle;

4. The radius of the inscribed circle ;

5. The radii of the escribed circles (See Art. 87).

Let ABC be the triangle; and let a, b, c denote the sides opposite the angles A, B, C respectively, and P1, P1, P, the perpendiculars drawn from the angles A, B, C ; ▲ the area of the triangle; R the radius of the circumscribing circle; r that of the inscribed circle; and r1, r r, the radii of the three escribed circles, which touch the sides a, b, c externally.

An escribed circle has already been defined (Art. 87) as a circle which touches one of the sides of a triangle exteriorly, and the other two sides produced.

We have already found (51) the perpendiculars to be

P1 √(a+b+c) (−a+b+c) (a−b+c) (a+b−c), [1]

P

=

=

2 a

√(a+b+c)(−a+b+c) (a−b+c) (a+b−c)

26

√(a+b+c) (−a+b+c)(a−b+c) (a+b−c)

2 c

[2]

[3]