The side lengths of a triangle are consecutive integers.
The largest interior angle is twice the smallest.

What are the side lengths of the triangle?

let the side lengths be x,x+1,x+2
let the smallest and largets angles be
a,2a

then from law of cosines we have
x^2=(x+1)^2+(x+2)^2-2(x+1)(x+2)cos(a)
x^2=x^2+2x+1+x^2+4x+4-2(x+1)(x+2)cos(a)
2(x+1)(x+2)cos(a)=x^2+6x+5
2(x+1)(x+2)cos(a)=(x+1)(x+5)
cos(a)=(x+5)/[2(x+2)]

we also have
(x+2)^2=x^2+(x+1)^2-2x(x+1)cos(2a)
x^2+4x+4=x^2+x^2+2x+1-2x(x+1)cos(2a)
2x(x+1)cos(2a)=x^2-2x-3
2x(x+1)cos(2a)=(x+1)(x-3)
cos(2a)=(x-3)/[2x]

thus
cos(2a)=2cos^2(a)-1=
[(x+5)^2/[2(x+2)^2]-1=
[(x+5)^2-2(x+2)^2]/[2(x+2)^2]=(x-3)/[2x]
x(x+5)^2-2x(x+2)^2=(x-3)(x+2)^2
x^3+10x^2+25x-2x^3-8x^2-8x=(x-3)(x^2+4x+4)
-x^3+2x^2+17x=x^3+x^2-8x-12
2x^3-x^2-25x-12=0
(x-4)(x+3)(2x+1)=0
x=4 x=-3 x=-1/2
only x=4 is valid for side length of triangle
sides are
4,5,6
angles are in degrees
41.40962211 55.77113367 82.81924422

 

Posted by Daniel on 2009-03-05 12:22:46