Question

Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the centre, angles of $\frac{\pi }{k}$ and $\frac{2\pi }{k}$, where $k>0$, then the value of $\left[k\right]$ is
[Note: $\left[k\right]$ denotes the largest integer less than or equal to $k$].

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is B $3$

Solution:

$2\cos \frac{\pi }{2k}+2\cos \frac{\pi }{k}=\sqrt{3}+1$

or, $\cos \frac{\pi }{2k}+\cos \frac{\pi }{k}=\frac{\sqrt{3}+1}{2}$

Let $\frac{\pi }{k}=\theta$

$\cos \frac{\theta }{2}+\cos \theta =\frac{\sqrt{3}+1}{2}$

$2co{s}^2\frac{\theta }{2}-1+\cos \frac{\theta }{2}=\frac{\sqrt{3}+1}{2}$

$\cos \frac{\theta }{2}=t$

or, $t^2+t-\frac{\sqrt{3}+3}{2}=0$

or, $t=\frac{-1\pm \sqrt{1+4(3+\sqrt{3})}}{4}=\frac{-1\pm (2\sqrt{3}+1)}{4}=\frac{-2-2\sqrt{3}}{4}$ or $\frac{\sqrt{3}}{2}$ $\because t\in [-1,1]$

or, $t=\cos \frac{\theta }{2}=\frac{\sqrt{3}}{2}$

or, $\frac{\theta }{2}=\frac{\pi }{6}⟹k=3$

Hence, B is the correct option.