Given the triangle ΔABC as shown below:
Required: value of x.
Step 1:
Using 30° as the focus angle, label the sides of the triangle.
Thus,
[tex]\begin{gathered} AB\Rightarrow hypotenuse \\ AC\Rightarrow adjacent \\ BC\Rightarrow opposite \end{gathered}[/tex]
Step 2:
Evaluate the length x
[tex]\begin{gathered} \tan \text{ 30 = }\frac{opposite}{adjacent} \\ \tan \text{ 30 = }\frac{BC}{AC} \\ \text{but tan 30 = }\frac{\sqrt{3}}{3} \\ \text{thus,} \\ \frac{\sqrt[]{3}}{3}\text{ = }\frac{\sqrt[]{2}}{x} \\ x\times\sqrt[]{3}\text{ = }\sqrt[]{2}\text{ }\times\text{ 3} \\ \Rightarrow x\text{ = }\frac{3\sqrt[]{2}\text{ }}{\sqrt[]{3}}\text{ } \end{gathered}[/tex]
Step 3:
Rationalize the denominator.
Thus,
[tex]\begin{gathered} \text{ }\frac{3\sqrt[]{2}\text{ }}{\sqrt[]{3}}\text{ }\times\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ =\frac{3\sqrt[]{2}\text{ }\times\sqrt[]{3}}{3} \\ =\frac{3\sqrt[]{6}\text{ }}{3} \\ \text{Thus, } \\ x=\sqrt[]{6}\text{ } \end{gathered}[/tex]
Hence, the lenth of
