Algebra 2 (1st Edition)

$x=y=\dfrac{\sqrt {42}}{2}$
Consider the lengths of the sides of the triangle, $x$ and $y$. $\sin \theta=\dfrac{Opposite}{Hypotenuse}=\dfrac{y}{h}$ and $\cos \theta=\dfrac{Adjacent}{Hypotenuse}=\dfrac{x}{h}$ Here, $\sin 45^{\circ}=\dfrac{y}{h} \implies \dfrac{1}{\sqrt 2}=\dfrac{y}{\sqrt {21}}$ This gives: $y=\dfrac{\sqrt {42}}{2}$ Now, $\cos 45^{\circ}=\dfrac{x}{h} \implies \dfrac{1}{\sqrt 2}=\dfrac{x}{\sqrt {21}}$ This gives: $x=\dfrac{\sqrt {42}}{2}$ Hence, $x=y=\dfrac{\sqrt {42}}{2}$