# Angle Formed by Three Squares +-- {.image} [[AngleFormedbyThreeSquares.png:pic]] > The squares are all the same size. What’s the angle? =-- ## Solution by [[Trigonometry]] +-- {.image} [[AngleFormedbyThreeSquaresLabelled.png:pic]] =-- With the points labelled as above, triangle $A B C$ is a [[right-angled triangle]] with $A C$ twice the length of $B C$ and so is half an [[equilateral triangle]]. This means that $G F C$ is also half an equilateral triangle, so if $x$ is the length of the side of one square, $C F$ has length $\frac{1}{2} x$ and $G F$ has length $\frac{\sqrt{3}}{2} x$. Then $E B$ has length $\frac{3}{2} x$ and $G E$ has length $\frac{\sqrt{3}}{2} x + x$. The tangent of angle $G \hat{D} E$ is therefore: $$ \frac{\frac{\sqrt{3}}{2}x + x}{\frac{3}{2} x} = \frac{2 + \sqrt{3}}{3} $$ So to $3$ decimal places, angle $I \hat{H} G$ is: $$ 180^\circ - \tan^{-1}\left(\frac{2 + \sqrt{3}}{3}\right) = 128.794^\circ $$