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

$\alpha = 45^\circ,\quad \beta = 120^\circ$

$\gamma = \theta\ (\text{acute, with the } z\text{-axis})$

$\text{To find: } \theta = \gamma \text{ (acute, with the } z\text{-axis)}$

$\text{Direction angles } \alpha,\ \beta,\ \gamma \text{ with the } x,\ y,\ z \text{ axes}$

$(l,m,n)=(\cos\alpha,\ \cos\beta,\ \cos\gamma)$

$l^2+m^2+n^2=1$

$\alpha=45^\circ \Rightarrow l=\cos 45^\circ=\dfrac{1}{\sqrt{2}}$

$\beta=120^\circ \Rightarrow m=\cos 120^\circ=-\dfrac{1}{2}$

$\text{Plan: } l,m \Rightarrow n^2 \Rightarrow \gamma \text{ (via the acute condition)}$

$l^2=\left(\dfrac{1}{\sqrt{2}}\right)^2=\dfrac{1}{2}$

$m^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}$

$l^2+m^2=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}$

$n^2=1-(l^2+m^2)=1-\dfrac{3}{4}=\dfrac{1}{4}$

$n=\pm\dfrac{1}{2} \Rightarrow \gamma=60^\circ \text{ or } 120^\circ$

$\text{acute} \Rightarrow \cos\gamma>0 \Rightarrow n=\dfrac{1}{2}$

$\cos\gamma=\dfrac{1}{2} \Rightarrow \gamma=60^\circ$

$\theta=60^\circ \quad \text{option }(2)$