Answer
$\angle A=45^{\circ}$ $;$ $\angle B=45^{\circ}$ $;$ $\angle C=90^{\circ}$
$a=16$ $;$ $b=16$ $;$ $h=16\sqrt{2}$
Work Step by Step
The triangle is shown below.
Let $h$ be the hypotenuse of the triangle, $a$ be the known cathetus and $b$ the unknown cathetus.
Two angles are known. Let $\angle A$ be the known angle and $\angle B$ and $\angle C$ the known ones. Since $\angle A+\angle B+\angle C=180^{\circ}$, substitute the known angles into the formula and solve for $\angle A$:
$\angle A=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$
Use the sine trigonometric ratio of the known angle to find $h$:
$\sin45^{\circ}=\dfrac{16}{h}$
$h=\dfrac{16}{\sin45^{\circ}}=\dfrac{16}{\Big(\dfrac{\sqrt{2}}{2}\Big)}=\dfrac{16(2)}{\sqrt{2}}=16\sqrt{2}$
Use the Pythagorean Theorem to obtain the unknown cathetus:
$b=\sqrt{(16\sqrt{2})^{2}-16^{2}}=\sqrt{(16^{2})(2)-16^{2}}=\sqrt{512-256}=...$
$...=\sqrt{256}=16$
