If the line x cos alpha + y sin alpha = 2 sqr | vedclass

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(B) The given line is $x \cos \alpha + y \sin \alpha = 2 \sqrt{3}$ and the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.Comparing with $\frac{x^2}{

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$\frac{\pi}{4}$

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(B) The given line is $x \cos \alpha + y \sin \alpha = 2 \sqrt{3}$ and the ellipse is $\frac{x^2}{16} + \frac{y^2}{8} = 1$.Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,we have $a^2 = 16$ and $b^2 = 8$.The condition for the line $x \cos \alpha + y \sin \alpha = p$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$.Here $p = 2 \sqrt{3}$,so $p^2 = (2 \sqrt{3})^2 = 12$.Substituting the values: $12 = 16 \cos^2 \alpha + 8 \sin^2 \alpha$.Using $\cos^2 \alpha = 1 - \sin^2 \alpha$,we get $12 = 16(1 - \sin^2 \alpha) + 8 \sin^2 \alpha$.$12 = 16 - 16 \sin^2 \alpha + 8 \sin^2 \alpha$.$8 \sin^2 \alpha = 4$.$\sin^2 \alpha = \frac{4}{8} = \frac{1}{2}$.Since $\alpha$ is an acute angle,$\sin \alpha = \frac{1}{\sqrt{2}}$.Therefore,$\alpha = \frac{\pi}{4}$.

If the line $x \cos \alpha + y \sin \alpha = 2 \sqrt{3}$ is a tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{8} = 1$ and $\alpha$ is an acute angle,then $\alpha = $

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