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(D) Let the direction angles of the line $AB$ be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.Given $\alpha = 45^{\circ}$ and $\beta =

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$\frac{\pi}{3}$

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(D) Let the direction angles of the line $AB$ be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.Given $\alpha = 45^{\circ}$ and $\beta = 120^{\circ}$.The relation between direction cosines is $\cos^{2} \alpha + \cos^{2} \beta + \cos^{2} \gamma = 1$.Substituting the values: $\cos^{2} 45^{\circ} + \cos^{2} 120^{\circ} + \cos^{2} \gamma = 1$.$\left(\frac{1}{\sqrt{2}}ight)^{2} + \left(-\frac{1}{2}ight)^{2} + \cos^{2} \gamma = 1$.$\frac{1}{2} + \frac{1}{4} + \cos^{2} \gamma = 1$.$\frac{3}{4} + \cos^{2} \gamma = 1$.$\cos^{2} \gamma = 1 - \frac{3}{4} = \frac{1}{4}$.Since $	heta = \gamma$ is an acute angle,$\cos \gamma = \frac{1}{2}$.Therefore,$\gamma = 60^{\circ} = \frac{\pi}{3}$.

$A$ line $AB$ in three-dimensional space makes angles $45^{\circ}$ and $120^{\circ}$ with the positive $x$-axis and the positive $y$-axis respectively. If $AB$ makes an acute angle $\theta$ with the positive $z$-axis,then $\theta$ equals

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