Let ABC be a triangle with A ( - 3, 1 ) and ∠ ACB = θ, 0 < θ <...

Let, $x -$coordinate of the point $C$ is $\alpha$, So coordinate of the point $C\left(\alpha, \dfrac{7\alpha - 1}{4}\right)$∵ ( C\text{lies on}7x - 4y - 1 = 0 )$</p><p> Hence, mid point of$AC$is$m\left(\dfrac{\alpha - 3}{2}, \dfrac{7\alpha + 3}{8}\right)$</p><p>As, point$m$lies on the line$2x + y - 3 = 0$, Substituting, we get$\alpha = 3$.</p><p>So.$C ( 3, 5 )$</p><p>Hence, slope of AC is$\dfrac{2}{3}$and slope of$7x - 4y - 1 = 0$is$\dfrac{7}{4}$</p><p>$∴ tan\dfrac{\theta}{2} = \dfrac{\dfrac{7}{4} - \dfrac{2}{3}}{1 + \dfrac{7}{4} \times \dfrac{2}{3}} = \dfrac{1}{2}$</p><p>$⇒ tan\theta = \dfrac{2 \cdot \dfrac{1}{2}}{1 - \dfrac{1}{4}} = \dfrac{4}{3} ⇒ tan\theta = \dfrac{4}{3}