What is the measure of the largest angle in a triangle with exradii 2, 3, and 4 ?

Given:
exradius_a = 2S/(-a + b + c) = 2
exradius_b = 2S/( a - b + c) = 3
exradius_c = 2S/( a + b - c) = 4

2S = 2(-a + b + c) = -2a + 2b + 2c
2S = 3( a - b + c) = 3a - 3b + 3c
2S = 4( a + b - c) = 4a + 4b - 4c

-2a + 2b + 2c = 3a - 3b + 3c
c = 5(b - a)

-2a + 2b + 2c = 4a + 4b - 4c
b = 3(c - a)

3a - 3b + 3c = 4a + 4b - 4c
a = 7(c - b)

a = 7(5(b - a) - b)
a = 35b - 35a - 7b
36a = 28b
a = 7/9 b AND b = 9/7 a

b = 3(c - 7/9 b)
b = 3c - 21/9 b
30/9 b = 3c
b = 9/10 c AND c = 10/9 b

c = 5(9/10 c - a)
c = 9/2c - 5a
7/2c = 5a
c = 10/7 a AND a = 7/10 b

Thus, if we let a = 7 then b = 9 and c = 10 and apply the law of cosines:

7² = 9² + 10² - 2*9*10*cos(A)
cos(A) = 11/15 A = ArcCos(11/15) ~= 0.747584 radians ~= 42.83343 degrees

9^2 = 7^2 + 10^2 - 2(7)(10)cos(B)
cos(B) = 67/140
B = ArcCos(67/140) ~= 1.071769 radians ~= 61.40786 degrees

10^2 = 7^2 + 9^2 - 2(7)(9)cos(C)
cos(C) = 5/21
C = ArcCos(5/21) ~= 1.330392 radians ~= 76.22585 degrees

The measure of the largest angle in a triangle with exradii 2, 3, 4 is, therefore, approximately equal to 1.330392 radians or 76.22585 degrees.

Posted by Dej Mar on 2007-11-15 14:38:48