The line 2x + y = 3 cuts the ellipse 4 x ^ 2 + y ^ 2 = 5 at points...

Solving $y = 3 - 2x$ and $4{x}^{2} + {y}^{2} = 5,$ we get the intersection points $$P\left(\dfrac{1}{2}, 2\right)$ and $Q\left(1,1\right)$
The equations of the tangents at these points are $4x\left(\dfrac{1}{2}\right) + 2y = 5$ and $4\left(x\right)\left(1\right) + y\left(1\right) = 5$
$⇒$ The slopes of the tangents at $P \text{ and } Q$ are $- 1 \text{ and } - 4$ respectively
$⇒$ The slopes of the normals at $P \text{ and } Q$ are $1 \text{ and } \dfrac{1}{4}$ respectively
$⇒ tan \theta = \left|\dfrac{1 - \dfrac{1}{4}}{1 + 1\left(\dfrac{1}{4}\right)}\right| = \dfrac{3}{5} ⇒ \theta = \tan^{- 1} \dfrac{3}{5}$
$⇒ cos \theta = \dfrac{5}{\sqrt{34}}, sin \theta = \dfrac{3}{\sqrt{34}} ⇒ \theta = \sin^{- 1} \dfrac{3}{\sqrt{34}}