Question

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

Answer 1

Hint: In this particular question first draw the pictorial representation of the above problem it will give us a clear picture of what we have to find out and use the concept that on any right angle triangle cosine is the ratio of base to hypotenuse, so use these concepts to reach the solution of the question.


Complete step-by-step answer:

Consider the circle with center O hand radius 2 units as shown in the above figure.
Therefore, OA = OC = OB = OD = 2 units

${AB}$ and ${CD}$ are two parallel chords that are at distance $(\sqrt{3}+1)$ units apart as shown above.

Now these chords subtend angles $\dfrac{\pi}{k}, \dfrac{2 \pi}{k}$ at the center as shown above. $\Rightarrow \angle A O B=\dfrac{\pi}{k}, \angle D O C=\dfrac{2 \pi}{k}$

Line ${EF}$ is passing through the center so angle AOE and angle ${COF}$ is the bisector of angle ${AOB}$ and angle

DOC respectively. $\Rightarrow \angle A O E=\dfrac{\angle A O B}{2}=\dfrac{\dfrac{\pi}{k}}{2}=\dfrac{\pi}{2 k}$

And

$\Rightarrow \angle C O F=\dfrac{\angle D O C}{2}=\dfrac{\dfrac{2 \pi}{k}}{2}=\dfrac{\pi}{k}$

Let ${OE}=\{a}$, therefore, ${OF}=(\sqrt{3}+1-a)$

Now in triangle AOE,

$\Rightarrow \cos \dfrac{\pi}{2 k}=\dfrac{\text { base }}{\text { hypotenuse }}=\dfrac{O E}{O A}=\dfrac{a}{2} \ldots \ldots \ldots \ldots . .$ (1)

And in triangle COF,

$\Rightarrow \cos \dfrac{\pi}{k}=\dfrac{\text { base }}{\text { hypotenuse }}=\dfrac{O F}{O C}=\dfrac{\sqrt{3}+1-a}{2} \ldots \ldots \ldots \ldots$

Now add equation (1) and (2) we have, $\Rightarrow \cos \dfrac{\pi}{2 k}+\cos \dfrac{\pi}{k}=\dfrac{a}{2}+\dfrac{\sqrt{3}+1-a}{2}$

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

Now let, $\dfrac{\pi}{2 k}=\theta, \Rightarrow \dfrac{\pi}{k}=2 \theta$ 

We have, 

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

Now as we know that, $\cos 2 \theta=2 \cos ^{2} \theta-1$

$\Rightarrow \cos \theta+2 \cos ^{2} \theta-1=\dfrac{\sqrt{3}+1}{2}$

Let, $\cos \theta={t}$

$\Rightarrow t+2 t^{2}-1=\dfrac{\sqrt{3}+1}{2}$

$\Rightarrow 4 t^{2}+2 t-2=\sqrt{3}+1$

$\Rightarrow 4 t^{2}+2 t-\sqrt{3}-3=0$

Now apply quadratic formula we have, $\Rightarrow t=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$, where, $a=4, b=2, c=(-3-\sqrt{3})$

$\Rightarrow t=\dfrac{-2 \pm \sqrt{2^{2}-4(4)(-3-\sqrt{3})}}{2(4)}$

$\Rightarrow t=\dfrac{-2 \pm \sqrt{4-16(-3-\sqrt{3})}}{8}$

$\Rightarrow t=\dfrac{-2 \pm \sqrt{52+16 \sqrt{3}}}{8}=\dfrac{-2 \pm 2 \sqrt{13+4 \sqrt{3}}}{8}=\dfrac{-1 \pm \sqrt{13+4 \sqrt{3}}}{4}=\dfrac{-1 \pm \sqrt{(2 \sqrt{3}+1)^{2}}}{4}$

$\Rightarrow t=\dfrac{-1+(2 \sqrt{3}+1)}{4}, \dfrac{-1-(2 \sqrt{3}+1)}{4}$

$\Rightarrow t=\dfrac{\sqrt{3}}{2}, \dfrac{-1-\sqrt{3}}{2}$

$\Rightarrow \cos \theta=\dfrac{\sqrt{3}}{2}, \dfrac{-1-\sqrt{3}}{2}$

As, $\dfrac{-1-\sqrt{3}}{2}=\dfrac{-1-1.732}{2}=\dfrac{-2.732}{2}=-1.366$

So, $\cos \theta=-1.366$ which is not possible as, $-1 \leq \cos \theta \leq 1$

So the possible case is, $\Rightarrow \cos \theta=\dfrac{\sqrt{3}}{2}=\cos \dfrac{\pi}{6}$

$\Rightarrow \theta=\dfrac{\pi}{6}$

But, $\dfrac{\pi}{2 k}=\theta$

$\Rightarrow \dfrac{\pi}{2 k}=\dfrac{\pi}{6}$

So on comparing, $2 \mathrm{k}=6$

Therefore, $k=3$

Now we have to find out the value of $k$, where $k$ denotes the greatest integer function less than or equal to $\mathrm{K}$.

Therefore,

$K=3=3$

So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the quadratic formula to solve the quadratic equation which is stated above and always recall that the greatest integer of x (say x = 0.1) i.e. [x] = [0.1] = 0 i.e. less than or equal to x.