# In a triangle ABC if A=pi/4 and tanBtanC=k then k must satisfy:​a) k^2 -6k +1 => 0b) k^2 -6k -1 = 0​c)  k^2-6k+1 d) 3 – 2root2

Riddhish Bhalodia
6 years ago
We know that in a triangle

and hence

Now is the tricky part
for the triangle to exist tanB and tanC must have a real solution
so now we have product ad sum of roots and thus we can generate a quadratic equation

whose roots are tanB and tanC
so now just making it’s determinant greater than equal to zero we get