Question

if A=pi/4 B=5pi/4 show a + c^1/2 = 2b

Answer 1

Answer:

Step-by-step explanation:

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

Answer 2

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