# Solution to the [[A Square in a Square]] Puzzle +-- {.image} [[ASquareinaSquare.jpeg:pic]] > The shaded square covers half the larger square. What’s the angle? =-- ## Solution by [[Area of a Triangle]] and Angles in an [[Equilateral Triangle]] +-- {.image} [[ASquareinaSquareAnnotated.jpeg:pic]] =-- In the above diagram, $D$ is the point where $D C$ and $C B$ have the same length, and $E$ is so that $D E$ is [[perpendicular]] to $A B$. Label the lengths in triangle $A C B$ by $a$, $b$, $c$ where $a$ is the length of $C B$, $b$ of $A C$, and $c$ of $A B$. Let $h$ be the length of $D E$. Triangle $A B C$ has area $\frac{1}{2} a b$, and four of them have half the area of the outer square, so $4 \times \frac{1}{2} a b = \frac{1}{2} c^2$, and thus $a b = \frac{1}{4} c^2$. Triangle $A D B$ has twice the area of triangle $A C B$, so has area $a b$. Its area is also $\frac{1}{2} c h$. Putting these together shows that $h = \frac{1}{2} c$. Since $A D$ is congruent to $A B$, triangle $A D E$ is a [[right-angled triangle]] with $D E$ half the length of $A D$, hence it is half of an [[equilateral triangle]] and so angle $D \hat{A} E$ is $30^\circ$, meaning that angle $C \hat{A} B$ is $15^\circ$. ## Solution by [[Trigonometry]] and the [[Double Angle Formula]] for Sine With the lengths as above, let $x$ be angle $C \hat{A} B$. Then: $$ \begin{aligned} \sin(x) &= \frac{a}{c} \\ \cos(x) &= \frac{b}{c} \end{aligned} $$ As above, $a b = \frac{1}{4} c^2$ so: $$ \frac{1}{2} = \frac{2 a b}{c^2 } = 2 \sin(x) \cos(x) = \sin(2 x) $$ Therefore $2 x = 30^\circ$ and so $x = 15^\circ$. $$