sect the sides A C, B C in L, M; and draw the lines L O, MO, perpendicular to them: O will be the centre of the circumscribed circle. (Euc. 111. 25.) Join AO, BO, CO, and let each be denoted by R; then,

4 √ S. (S—a). (S—b) • (S−c)

a b c

Hence, also R. r =

=

4 S

a b c
2 (a+b+c)

(65.) To find the area of the quadrilateral figure that can be inscribed in a circle.

Let the sides AB, BC, C D, AD (fig. 17.) be

denoted by a, b, c, d : also let the angle A B C=0, ... AD C=π-0 (Euc. 111. 22.) Join A C.

The area of the quadrilateral A B C D

Now, by (49.), a2+b2 −2 a b cos 0= A C2 =c2+d2-2 c d cos (π-0)=c2+d2+2 c d cos 0;

_(c+d)2—(a—b)2 (a+b)2—(c—d)2
2 (ab+cd)

2 (a b+cd)

(b+c+d—a)·(a+c+d—b). (a+b+d−c). (a+b+c−d);

4 (a b + c d )3

suppose a+b+c+d=2 S'; and we then have,

sin2 0:

4 (S'-a). (S'—b) . (S'—c) · (S'—d) (a b+cd)2

hence, sin

2

Fab+cd

✔ (S'—a) . (S'—b) . ($′′′—c) . (S′′—d) ;

therefore, by substitution in the foregoing expression for the area we obtain,

area A B C D

=√ (S'—a). (S'—b) . (S'—c) · (S—d).

The value of the diagonal A C expressed in terms of the four sides may readily be obtained.

We have, A C—a2+b2−2 a b cos 0

(66.) To find the area of a regular polygon. Let A and B (fig. 18.) be two contiguous angles of the polygon, and let them be bisected by the straight lines AO, BO; then the polygon will evidently be composed of n triangles, each equal to

AOB; also let AB-a, and draw OP perpen

log area log n-log 4+2 log a+log cot-10.

n

Example. Let a=1, n=7, then=25°42′51′′3,

hence, log area of heptagon

n

=log 1·75+log cot 25° 42′ 51′′3-10.

log 1.75

....

0.2430380

log cot 25° 42′ 51"3..... 10-3173364

log area..

0.5603744

... Area 3.633912.

(67.) We shall conclude this section with the following examples for exercise in plane trigonometry.

1. At the distance of 170 feet from the bottom of a tower, the angle of elevation of the top was observed with a quadrant to be 52° 30′. What was the height of the tower?

Answer, 221-55 feet.

2. Required the perpendicular height of a hill, the angle of elevation taken at the bottom of it being 46°, and 200 yards further off, on a level with the bottom, the angle was 31°.

Answer, 286 28 yards.

3. In any right angled triangle ABC, C being the right angle,

4. What was the perpendicular height of a balloon, when the angles of elevation, taken by two observers at the same time on opposite sides of it, and in the same vertical plane, were 65° 55′ and 71° 11′; the observers being separated 470 yards?

Answer, 596 655 yards.

5. The two sides of a plane triangle are 71848