Let the angles made with the positive x -axis by two straight lines drawn from the point $P(2,3)$ an

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Let the angles made with the positive x -axis by two straight lines drawn from the point $P(2,3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point P be $	heta_{1}$ and $	heta_{2}$. Then the value of $\left( 	heta_{1} + 	heta_{2} ight)$ is :

Accepted Answer

Let Q is $\left( \sqrt{\frac{2}{3}}cos	heta + 2,\sqrt{\frac{2}{3}}sin	heta + 3 ight)$so, $x + y = 6$$${\sqrt{\frac{2}{3}}(cos	heta + sin	heta) + 5 = 6 }{sin	heta + cos	heta = \sqrt{\frac{3}{2}} }{1 + sin2	heta = \frac{3}{2} }{sin2	heta = \frac{1}{2} }{2	heta = \frac{\pi}{6},\frac{5\pi}{6} }{	heta = \frac{\pi}{12}\&\frac{5\pi}{6} }$$So $	heta_{1} + 	heta_{2} = \frac{\pi}{2}$

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