Question

The angle between the straight lines $x^2+4xy+y^2=0$ is $\theta$ then $\sin \theta =$

• A. $\frac{1}{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $1$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (D)

$\frac{\sqrt{3}}{2}$

Given $x^2+4xy+y^2=0$

Here, $a=1,h=2,b=1$

$tan\theta =\frac{2\sqrt{h^2-ab}}{a+b}$

$tan\theta =\frac{2\sqrt{4-1}}{2}$

$tan\theta =\sqrt{3}$

$\therefore \theta =\frac{\pi }{3}$

$\therefore sin\theta =\frac{\sqrt{3}}{2}$