To helps us solve the problem we will give names to the missing lengths and the triangles:
We know that in any 30-60-90 triangle we have that:
• Side opposite the 30° angle: x
,
• Side opposite the 60° angle: x√3
,
• Side opposite the 90° angle: 2x
In triangle I we know the side opposite to the 60° angle and its value is 5, from the theorem and the names we gave we have that:
[tex]\begin{gathered} w\sqrt[]{3}=5 \\ w=\frac{5}{\sqrt[]{3}} \\ w=\frac{5\sqrt[]{3}}{3} \end{gathered}[/tex]
Once we know this we can use the theorem to find y which is the hypotenuse (opposite side to the 90° angle) and that, in this case, has to be twice w, then:
[tex]\begin{gathered} y=2w \\ y=2\cdot\frac{5\sqrt[]{3}}{3} \\ y=\frac{10\sqrt[]{3}}{3} \end{gathered}[/tex]
Now, for triangle II we once again have the opposite side to the 60° angle; using the theorem we have that:
[tex]\begin{gathered} x\sqrt[]{3}=\frac{10\sqrt[]{3}}{3} \\ x=\frac{10\sqrt[]{3}}{3\sqrt[]{3}} \\ x=\frac{10}{3} \end{gathered}[/tex]
Finally we know that z has to be twice x, then:
[tex]\begin{gathered} z=2x \\ z=2\cdot\frac{10}{3} \\ z=\frac{20}{3} \end{gathered}[/tex]
Therefore, summing up, we have that:
[tex]\begin{gathered} w=\frac{5\sqrt[]{3}}{3} \\ y=\frac{10\sqrt[]{3}}{3} \\ x=\frac{10}{3} \\ z=\frac{20}{3} \end{gathered}[/tex]
