Problem source

A   pyramid stands on horizontal ground. Its base is an equilateral triangle with sides of length~$a$,  the other three 
 sides of the pyramid are of length $b$ and its volume is $V$. Given that the 
formula for the volume of any pyramid is   
$
 \textstyle
\frac13 \times \mbox{area of base} \times \mbox {height} \,,
$
show that 
\[
V= \frac1{12} {a^2(3b^2-a^2)}^{\frac12}\;.
\]
The pyramid is then placed  so that a non-equilateral face  lies on the ground.
Show that the new height, $h$, of the pyramid is given by 
\[
h^2 = \frac{a^2(3b^2-a^2)}{4b^2-a^2}\;.
\]
Find, in terms of $a$ and $b\,$, the angle between the 
equilateral triangle and the horizontal.

Solution source

First let's consider the area of the base. It is an equilateral triangle with side length $a$, so $\frac12 a^2 \sin 60^\circ = \frac{\sqrt{3}}4a^2$.

\begin{center}
    \begin{tikzpicture}
        \draw (2,0) -- (0,0) node[pos=0.5, below] {distance to centre};
        \draw (0,0) -- (2, 4) node[pos=0.5, left] {$b$};
        \draw[dashed] (2,0) -- (2,4) node[pos=0.5, right] {$h$};
    \end{tikzpicture}
\end{center}

Let's consider the height. The distance to the centre $\frac23 \frac{\sqrt{3}}2 a = \frac{a}{\sqrt{3}}$ so $h = \sqrt{b^2 - \frac{a^2}{3}}$ and therefore the volume is:

\begin{align*}
V &= \frac13 \times \mbox{area of base} \times \mbox {height} \\
&= \frac13 \frac{\sqrt{3}}{4}a^2 \sqrt{\frac{3b^2-a^2}{3}} \\
&= \frac1{12}a^2 (3b^2-a^2)^{\frac12}
\end{align*}

The area of an isoceles triangle with sides $a,b,b$ can be found by considering the perpendicular:

\begin{center}
    \begin{tikzpicture}
        \draw (2,0) -- (0,0) node[pos=0.5, below] {$\frac{a}2$};
        \draw (0,0) -- (2, 4) node[pos=0.5, left] {$b$};
        \draw[dashed] (2,0) -- (2,4) node[pos=0.5, right] {$h$};
    \end{tikzpicture}
\end{center}

ie $\frac{a}{4} \sqrt{b^2-\frac{a^2}{4}} = \frac{a\sqrt{4b^2-a^2}}{8}$.

Therefore by considering the volume, we must have

\begin{align*}
&& V &= \frac13 \times \mbox{area of base} \times \mbox {height} \\
\Rightarrow && \frac1{12}a^2 (3b^2-a^2)^{\frac12} &= \frac13  \frac{a\sqrt{4b^2-a^2}}{8} h \\
\Rightarrow && h &= \frac{2a(3b^2-a^2)}{(4b^2-a^2)^{\frac12}} \\
\Rightarrow && h^2 &= \frac{4a^2(3b^2-a^2)}{4b^2-a^2}
\end{align*}