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(B) Let the direction cosines of the line be $l, m, n$. Since the line makes the same angle $\alpha$ with the $x$ and $y$ axes,we have $l = \cos \alph

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$\frac{\pi}{4}$

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(B) Let the direction cosines of the line be $l, m, n$. Since the line makes the same angle $\alpha$ with the $x$ and $y$ axes,we have $l = \cos \alpha$ and $m = \cos \alpha$. Let the angle with the $z$-axis be $	heta$,so $n = \cos 	heta$.We know that $l^2 + m^2 + n^2 = 1$,which implies $\cos^2 \alpha + \cos^2 \alpha + \cos^2 	heta = 1$.Using the identity $\cos^2 \phi = 1 - \sin^2 \phi$,we get $(1 - \sin^2 \alpha) + (1 - \sin^2 \alpha) + (1 - \sin^2 	heta) = 1$.$3 - 2 \sin^2 \alpha - \sin^2 	heta = 1$.Given $\sin^2 	heta = 2 \sin^2 \alpha$,substituting this into the equation:$3 - 2 \sin^2 \alpha - 2 \sin^2 \alpha = 1$.$3 - 4 \sin^2 \alpha = 1$.$4 \sin^2 \alpha = 2$.$\sin^2 \alpha = \frac{1}{2} = \sin^2 \frac{\pi}{4}$.Thus,$\alpha = \frac{\pi}{4}$.

$A$ line makes the same angle $\alpha$ with each of the $x$ and $y$ axes. If the angle $\theta$,which it makes with the $z$-axis,is such that $\sin^2 \theta = 2 \sin^2 \alpha$,then the angle $\alpha$ is

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