Solution
- The way to solve the question is to use SOHCAHTOA.
- This is given below:
[tex]\begin{gathered} \text{ We first need to find the Hypotenuse of the 45-45-90 triangle.} \\ \frac{11}{Hypotenuse}=\sin45 \\ \\ \therefore Hypotenuse=\frac{11}{\sin45}=11\sqrt{2} \\ \\ \\ \text{ The Hypotenuse of the 45-45-90 triangle is the Opposite of the 30-60-90 triangle} \\ \text{ Thus, we can find }x\text{ as follows:} \\ \\ \frac{11\sqrt{2}}{x}=\cos60 \\ \\ \frac{11\sqrt{2}}{\cos60}=x \\ \\ x=\frac{11\sqrt{2}}{\frac{1}{2}}=2\times11\sqrt{2} \\ \\ x=22\sqrt{2} \end{gathered}[/tex]
- Thus, the answer is
[tex]x=22\sqrt{2}[/tex]
