Question

Find the missing side lengths. Leave your answers as radicals in simplest form. I’m so confused!!!! Number 13

Answer 1

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:

`\begin{gathered} w\sqrt[]{3}=5 \\ w=\frac{5}{\sqrt[]{3}} \\ w=\frac{5\sqrt[]{3}}{3} \end{gathered}`

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:

`\begin{gathered} y=2w \\ y=2\cdot\frac{5\sqrt[]{3}}{3} \\ y=\frac{10\sqrt[]{3}}{3} \end{gathered}`

Now, for triangle II we once again have the opposite side to the 60° angle; using the theorem we have that:

`\begin{gathered} x\sqrt[]{3}=\frac{10\sqrt[]{3}}{3} \\ x=\frac{10\sqrt[]{3}}{3\sqrt[]{3}} \\ x=(10)/(3) \end{gathered}`

Finally we know that z has to be twice x, then:

`\begin{gathered} z=2x \\ z=2\cdot(10)/(3) \\ z=(20)/(3) \end{gathered}`

Therefore, summing up, we have that:

`\begin{gathered} w=\frac{5\sqrt[]{3}}{3} \\ y=\frac{10\sqrt[]{3}}{3} \\ x=(10)/(3) \\ z=(20)/(3) \end{gathered}`