Find the number of of all integer sided isosceles obtuse angle triangles with perimeter 2008.
The Attempt at a Solution
I assumed the obtuse angle as 90+2a
Let the length of the equal sides be L an that of the base = 2X
Draw the perpendicular to the side opposite to the obtuse angle from the opposite vertex.
It will divide the triangle in to congruent parts. Using sine rule in one of the to triangles formed.
X=1004/ (cosec (45+a)+1)
as 2X is an integer
cosec (45+a)+1 should be an integer and a factor of 1004 which is not possible. Therefore, no such triangle is possible!