109. Examples.

1. The sides of a triangle are 7, 9, 10, respectively, required the radius of the circumscribed circle.

Ans. 5.148.

2. The sides of a triangle are 50, 60, 70, respectively, required the radius of the circumscribed circle.

Ans. 35.72.

110. Theorem.

The perpendicular let fall on either side of a triangle from the vertex of the opposite angle is equal to that side into the product of the sines of the adjacent angles divided by the sine of the sum of those angles.

Given the three sides of a triangle to find the radii of the escribed circles.

The escribed circles are the three circles external to the triangle, each tangent to one side and to the prolongation of the other sides.

The centers of the escribed circles are the points of

Substituting the value of tan 4, article 98, we have

1. Given the sides of a triangle, 6, 9, 11, required the radii of the three escribed circles.

Ans. 3.854, 6.745, 13.49.

2. Given p=100, A=55°, B = 60°, C= 65°, required

the radii of the three escribed circles.

[See (2), Art. 111.]

Ans. 26.028, 28.867, 31.854.

113. Theorem.

The product of the radius of the inscribed circle and the radii of the three escribed circles is equal to the square of the area of the triangle.

The product of (8), article 106, and (3), (4), (5), article 111, gives

The reciprocal of the radius of the inscribed circle, the sum of the reciprocals of the radii of the escribed circles, and the sum of the reciprocals of the perpendiculars let fall from the vertices of the three angles on the opposite sides of a triangle are equal to each other.

Taking the reciprocal of (8), article 106, we have

Taking the sum of the reciprocals of (3), (4), (5),

Let p', p", p"", respectively, be the perpendiculars let fall from the vertices of the three angles on the sides a, b, and c. Then we have

To find the distance between the centers of the circumscribed and inscribed circles of a triangle.

Let R and r be the radii, and P and O the centers of the circles, and let D

=

OP.

Draw PE perpendicular to AC. The A

angle APE =

=

B, since each is meas

B

ured by one-half the arc AC; but PAE 90° — APE, ... PAE 90°-B. OAC A. PAO-PAE-OAC.

... PAO 90° - B-1A (C-B). A0=

r

sin А

(1) OP2=AP2+A02-2 APXAO cos PAO. Art. 97. Substituting the values of OP, AP, AO, and PAO, we have

Substituting in (2), and reducing by article 91, (d),

1. The sides of a triangle are 12, 13, 15; required the distance between the centers of the circumscribed and inscribed circles. Ans. 1.616.

2. Two sides of a triangle are 35 and 37, and their included angle is 50°; required the distance between the centers of the circumscribed and inscribed circles. Ans. 3.266.

3. The perimeter of a triangle is 120, the angles are 40°, 60°, and 80', respectively; required the distance between the centers of the circumscribed and inscribed circles. Ans. 8.353.

117. Problem.

To find the distance between the centers of the circumscribed and escribed circles.

Let r', r", r'" be the radii of the escribed circles, and D', D', D"", be the distances of their centers, O', 0", O"", respectively, from P, the center of the circumscribed circle, whose radius is R.

O"

0"