Posted by dat on Sunday, May 8, 2011 at 11:56am.
let the other two sides be a and b
then a^2 + b^2 = 100
the cotangents of the two acute angles are
a/b + b/a and of course the cot 90° = 0
a/b + b/a + 0 = 2
then a^2 + b^2/ab = 2
a^2 + b^2 = 2ab
a^2 - 2ab + b^2 = 0
(a - b)^2 = 0
a-b = 0
a = b
sub back into a^2 + b^2 = 100
a^2 + a^2 = 100 , since a=b
a^2 = 50
a = 5√2
also if a^2 = 50
then b^2 = 50 and b= 5√2
It must be an isosceles right-angled triangle with the other sides 5√2 each
check: tan (of angle) = (5√2/5√2) = 1
so 1 + 1 + 0 = 2
Is (5√2)^2 + (5√2)^2 = 100 ? yes!
is your "p" supposed to say √ ??
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