Solid Trigonometry

Question

V ABCD is a pyramid with the vertex V situated perpendicularly above the centre of the square base ABCD. If $\theta$ is the angle between the edge VA and the base, and $\phi$ is the angle between the plane VAB and the base, show that tan$\phi$=$\sqrt{2}$tan$\theta$.

Answer 1

Let $X$ be the center of the base of the given pyramid, and let $M$ be the midpoint of the edge $AB$. The line $XM$ is a translate of $DA$, so it is perpedicular to $AB$. Also, $VM$ is the bisector of the angle $V$ in the isosceles triangle $AVB$, using the fact that the base angles of an isosceles triangle are equal you get $VM$ is perpendicular to $AB$. It follows that $\phi$ is the measure of the angle $M$ in the triangle $VMC$. Now, since $VX$ is perpendicular to the base of the pyramid, $\theta$ is the angle at $A$ in $VAX$. Since $AX=\sqrt2\cdot MX$, $$\tan\theta=\frac{VX}{AX}=\frac1{\sqrt2}\cdot\frac{VX}{MX}=\frac1{\sqrt2}\tan\phi$$