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

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

Grade:11

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
6 years ago
We know that in a triangle
tanA+tanB+tanC = tanAtanBtanC
and hence
tanB+tanC = k-1
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
f(x) = x^2 - (k-1)x +k =0
whose roots are tanB and tanC
so now just making it’s determinant greater than equal to zero we get
(k-1)^2 - 4k = k^2-6k+1 \geq 0
So answer is option A

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