Step-by-step explanation:
Using trigonometrical functions we can obtain the required side lengths.
[tex] \sin 45\degree = \frac{a}{16\sqrt 2}\\\\
\therefore \frac{1}{\sqrt 2}= \frac{a}{16\sqrt 2}\\\\
\therefore a = \frac{16\sqrt 2}{\sqrt 2}\\\\
\huge\red {\boxed {\therefore a = 16}} \\\\
\cos 45\degree = \frac{c}{16\sqrt 2}\\\\
\therefore \frac{1}{\sqrt 2}= \frac{c}{16\sqrt 2}\\\\
\therefore c = \frac{16\sqrt 2}{\sqrt 2}\\\\
\huge\purple {\boxed {\therefore c = 16}} \\\\
\sin 30\degree = \frac{a}{b}\\\\
\therefore \frac{1}{2}= \frac{16}{b}\\\\
\therefore b = {16\times2}\\\\
\huge\orange{\boxed {\therefore b = 32}} \\\\
\tan 30\degree = \frac{a}{d}\\\\
\therefore \frac{1}{\sqrt 3}= \frac{16}{d}\\\\
\therefore d = {16\times\sqrt 3}\\\\
\huge\pink {\boxed {\therefore d = 16\sqrt 3}} \\\\
[/tex]
