trig
posted by dat .
In a right triangle the hypotenuse has length 10 and the the sum of the cotangents of
all three angles of the triangle equals 2. What are the lengths of the other two sides
of the triangle?
(A) 5 and 5p3
(B) 5p2 and 5p2
(C) 1 and3
p11 (D) 4 and 2p21
(E) None of the above.

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
ab = 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 rightangled 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 √ ??