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

For any directed line in 3-D, let $$\alpha ,\beta ,\gamma$$ be the angles it makes with the positive $$x,y,z$$ axes respectively. Its direction cosines are $$l=\cos\alpha,\; m=\cos\beta,\; n=\cos\gamma$$ and they always satisfy

$$l^{2}+m^{2}+n^{2}=1 \qquad -(1)$$

Here the given angles are $$\alpha =45^\circ,\; \beta =120^\circ,\; \gamma=\theta\;(\theta \text{ acute})$$

Compute the known direction cosines:

$$l=\cos45^\circ=\frac{1}{\sqrt2}, \qquad m=\cos120^\circ=-\frac12$$

Substitute $$l,m$$ into equation $$-(1)$$:

$$\left(\frac{1}{\sqrt2}\right)^{2}+\left(-\frac12\right)^{2}+n^{2}=1$$ $$\frac12+\frac14+n^{2}=1$$ $$\frac34+n^{2}=1$$ $$n^{2}=\frac14$$

Since the required angle with the positive $$z$$-axis is acute, $$n=\cos\theta$$ must be positive, so $$n=\frac12$$.

Therefore

$$\theta=\cos^{-1}\!\left(\frac12\right)=60^\circ$$

Hence the line makes an acute angle of $$60^\circ$$ with the positive $$z$$-axis.

Option B which is: $$60^\circ$$