To find z, consider the right-angled triangle at the botton in the diagram showm
[tex]\begin{gathered} \sin 45\text{ = }\frac{z}{20} \\ z\text{ = 20 }\sin 45 \\ z\text{ = }20\text{ }\times\frac{1}{\sqrt[]{2}} \\ z\text{ = }\frac{20}{\sqrt[]{2}} \\ z\text{ = }\frac{20}{\sqrt[]{2}}\times\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ z\text{ = }\frac{20\sqrt[]{2}}{2} \\ z\text{ = 10}\sqrt[]{2} \end{gathered}[/tex]
Let the common base of both triangles be m
[tex]\begin{gathered} \cos 45\text{ = }\frac{m}{20} \\ m\text{ = 20 }\cos 45 \\ m\text{ = }\frac{20}{\sqrt[]{2}} \\ m\text{ = }\frac{20}{\sqrt[]{2}}\times\frac{\sqrt[]{2}}{\sqrt[]{2}} \\ m\text{ = 10}\sqrt[]{2} \end{gathered}[/tex]
To find y:
[tex]\begin{gathered} \tan 30\text{ = }\frac{y}{m} \\ \tan 30\text{ = }\frac{y}{10\sqrt[]{2}} \\ \frac{1}{\sqrt[]{3}}=\text{ }\frac{y}{10\sqrt[]{2}} \\ y\text{ = }\frac{10\sqrt[]{2}}{\sqrt[]{3}} \\ y\text{ = }\frac{10\sqrt[]{6}}{3} \end{gathered}[/tex]
To find x:
[tex]\begin{gathered} \sin 30=\frac{y}{x} \\ \frac{1}{2}=\frac{10\sqrt[]{6}}{3}\div x \\ \frac{1}{2}=\frac{10\sqrt[]{6}}{3}\times\frac{1}{x} \\ x\text{ = }\frac{20\sqrt[]{6}}{3} \end{gathered}[/tex]
