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(C) The equation of the line is $2x + 7y + 6 = 0$.Slope of the line $m = -\frac{2}{7}$.The normal drawn from the origin to the line is perpendicular t

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$\pi + 	an^{-1} \frac{7}{2}$

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(C) The equation of the line is $2x + 7y + 6 = 0$.Slope of the line $m = -\frac{2}{7}$.The normal drawn from the origin to the line is perpendicular to the line.Let the slope of the normal be $m'$. Since the normal is perpendicular to the line,$m 	imes m' = -1$.$(-\frac{2}{7}) 	imes m' = -1 \Rightarrow m' = \frac{7}{2}$.Let $\alpha$ be the angle the normal makes with the positive $X$-axis. Then $	an \alpha = m' = \frac{7}{2}$,so $\alpha = 	an^{-1} \frac{7}{2}$.From the figure,the normal lies in the third quadrant,so the angle $	heta$ with the positive $X$-axis is $\pi + \alpha$.Therefore,$	heta = \pi + 	an^{-1} \frac{7}{2}$.

If the normal drawn from the origin to the straight line $2x + 7y + 6 = 0$ makes an angle $\theta$ with the positive $X$-axis,then $\theta =$

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