Question

7. If the length of the longer leg of a 30-60-90 triangle is x V3, then thelength of the hypotenuse isОх2xXV2O x73

Answer 1

Let's illustrate this with a diagram

Now we will use SOHCAHTOA to solve this

We have just the adjacent side and the hypotenuse

`\begin{gathered} \sin 60^o=\frac{opposite}{\text{hypotenuse }} \\ \\ \sin 60^o=\frac{x\sqrt[]{3}}{y} \\ \\ y\sin 60^o=x\sqrt[]{3} \\ y\text{ = }\frac{x\sqrt[]{3}}{\frac{\sqrt[]{3}}{2}} \\ \\ y=x\sqrt[]{3}*\frac{2}{\sqrt[]{3}} \\ \\ \sqrt[]{3}\text{ cancels out} \\ \\ y=x*2 \\ y=2x \end{gathered}`

Therefore, the hypotenuse is 2x