A line makes angles 60^{circ}, 45^{circ}, the | vedclass

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(A) Let the direction angles of the line be $\alpha = 60^{\circ}$, $\beta = 45^{\circ}$, and $\gamma = 	heta$.We know that the sum of the squares of

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$\sqrt{3}$

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(A) Let the direction angles of the line be $\alpha = 60^{\circ}$, $\beta = 45^{\circ}$, and $\gamma = 	heta$.We know that the sum of the squares of the direction cosines of a line is $1$, which is given by the formula $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.Substituting the given values, we get $\cos^2 60^{\circ} + \cos^2 45^{\circ} + \cos^2 	heta = 1$.Since $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$, we have $(\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 	heta = 1$.$\frac{1}{4} + \frac{1}{2} + \cos^2 	heta = 1$.$\frac{3}{4} + \cos^2 	heta = 1$.$\cos^2 	heta = 1 - \frac{3}{4} = \frac{1}{4}$.Since $	heta$ is an acute angle, $\cos 	heta = \frac{1}{2}$.Thus, $	heta = 60^{\circ}$.Therefore, $	an 	heta = 	an 60^{\circ} = \sqrt{3}$.

$A$ line makes angles $60^{\circ}, 45^{\circ}, \theta$ with positive $X, Y, Z$-axes respectively. If $\theta$ is an acute angle, then $\tan \theta=$

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