Solid Geometry – Problem 14: Find the angle between the space diagonal and the base plane

In a right triangular prism $ABC-A_1B_1C_1$, the base $ABC$ is an equilateral triangle with side $2$, and $AA_1=BB_1=CC_1=3$. Find the angle between the space diagonal $AC_1$ and the base plane $ABC$.

Build a coordinate system with plane $ABC$ as $z=0$. Take $$A(0,0,0),\ B(2,0,0),\ C(1,\sqrt3,0),\ C_1(1,\sqrt3,3).$$ Then a direction vector of $AC_1$ is $$\overrightarrow{AC_1}=(1,\sqrt3,3).$$ Its projection on the base plane $z=0$ is $(1,\sqrt3,0)$, whose length is $$\sqrt{1^2+(\sqrt3)^2}=2.$$ The length of $\overrightarrow{AC_1}$ is $$|\overrightarrow{AC_1}|=\sqrt{1^2+(\sqrt3)^2+3^2}=\sqrt{13}.$$ Let $\theta$ be the angle between line $AC_1$ and plane $ABC$. Then $$\sin\theta=\frac{\text{(perpendicular component)}}{|\overrightarrow{AC_1}|}=\frac{3}{\sqrt{13}},$$ and equivalently $$\tan\theta=\frac{3}{2}.$$

If $\theta$ is the angle between $AC_1$ and plane $ABC$, then $\sin\theta=\dfrac{3}{\sqrt{13}}$ (so $\theta=\arctan\dfrac{3}{2}$).