The acute angle of a rhombus whose side is a mean proportion al between its diagonals is

Let the length of the diagonals of the rhombus be $a$ and $b$ and $x$ be its side then

$\begin{array}{l}\frac{x}{a}=\frac{b}{x}\Rightarrow x^2=ab\\ \text{ Now }\cos \theta =\frac{x^2+x^2-a^2}{2x\cdot x}=\frac{2ab-a^2}{2ab}=1-\frac{a}{2b}\\ \text{ Similarly }\cos \left({180}^{\circ }-\theta \right)=1-\frac{b}{2a}\\ \Rightarrow \frac{a}{2b}=1-\cos \theta ,\frac{b}{2a}=1+\cos \theta \\ \Rightarrow 1-{\cos }^2\theta =\frac{1}{4}\Rightarrow \sin \theta =\frac{1}{2}\Rightarrow \theta ={30}^{\circ }\end{array}$