Finding values of $\theta$ for which $(\cos\theta,\sin\theta)$ lies inside the triangle formed by $x+y=2$, $x−y=1$, and $6x+2y−\sqrt{10}​=0$

Find the range of values of $$\theta$$, such that $$\theta\in[0,2\pi]$$ for which $$(\cos \theta,\sin \theta)$$ lies inside the triangle formed by $$x + y = 2, x − y = 1$$ and $$6x + 2y − \sqrt{10} ​=0$$.

Not getting any hint of how can I get the range of those points is there any short method of doing so.

I have plotted the graph and then I found out that the line $$6x + 2y − \sqrt{10} ​=0$$ which forms one of the sides of the triangle has $$O(0,0)$$ and $$P(\cos \theta, \sin\theta)$$ on opposite sides so applied the formula of power of point.

So $$L_3(O)\cdot L_3(P)<0$$ and then solved to find out the range.

If there is any short method please tell me. • Maybe the people who voted to close could leave a helpful comment saying what was unclear? There is a triangle, and there is a circle, and the question is asking about where they meet - what's unclear? Or am I missing something? May 8 '21 at 10:56
• (p.s. @Vaibhav you should use MathJax to make your question more readable.) May 8 '21 at 10:57
• sir, but it seems clear to me , I don't understand what is not clear, people are just downvoting my question as they are unable to understand the problem, its not that easy problem its from IIT jee how would anyone without enough knowledge solve it May 8 '21 at 22:42
• Thanks for helping Sir, I have posted that request. May 12 '21 at 19:12
• Do you recognize what you get when you look at all the points $(\cos\theta,\sin\theta)$ for $0\le\theta\le2\pi$, Vaibhav? Then you just need to find where some lines intersect that very special curve. I don't know just why users voted down, but I'm sure they had the knowledge to solve it. May 12 '21 at 22:00

Working from the diagram OP has posted, $$D=(1,0)$$ corresponds to $$\theta=0$$, so we only need to work out $$E$$. Now, $$E$$ is the intersection of $$6x+2y=\sqrt{10}$$ and $$x^2+y^2=1$$. In polar coordinates, these are $$6r\cos\theta+2r\sin\theta=\sqrt{10}$$ and $$r=1$$, so it comes down to solving $$6\cos\theta+2\sin\theta=\sqrt{10}$$. Noting that $$6^2+2^2=40$$, we divide both sides by $$\sqrt{40}=2\sqrt{10}$$ to get $${3\over\sqrt{10}}\cos\theta+{1\over\sqrt{10}}\sin\theta={1\over2}$$ Let $$\eta=\arcsin{3\over\sqrt{10}}$$ which implies $$\cos\eta={1\over\sqrt{10}}$$ and $$\tan\eta=3$$. Then we have $$1/2=\sin\eta\cos\theta+\cos\eta\sin\theta=\sin(\theta+\eta)$$. This is satisfied by $$\theta+\eta=\pi/6$$, but $$\tan\eta=3>1$$ implies $$\eta>\pi/4$$ and thus $$\theta+\eta>\pi/4$$, so we go to $$\theta+\eta=5\pi/6$$. Then, $$\theta={5\pi\over6}-\eta={5\pi\over6}-\arctan3$$