First, let's remember the exact value of the sine and cosine of 45°. If we have a right triangle with sides 1 and hypotenuse sqrt(2), then we have:
Then, we have the following values for the sine and cosine of 45°:
![\begin{gathered} \sin (45)=\frac{\text{opposite side}}{hypotenuse}=\frac{1}{\sqrt[]{2}} \\ \cos (45)=\frac{\text{adjacent side}}{hypotenuse}=\frac{1}{\sqrt[]{2}} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/v0cngkev8abfvpp5kw4bcp2kyi1g5raarp.png)
We can find the values of x and y using the trigonometric functions sine and cosine:
![\begin{gathered} \sin (45)=\frac{\text{opposite side}}{hypotenuse} \\ \Rightarrow\sin (45)=(x)/(24) \\ \Rightarrow\frac{1}{\sqrt[]{2}}\cdot24=x \\ \Rightarrow x=(\frac{24}{\sqrt[]{2}})\cdot(\frac{\sqrt[]{2}}{\sqrt[]{2}})=\frac{24\cdot\sqrt[]{2}}{2}=12\sqrt[]{2} \\ \cos (45)=\frac{\text{adjacent side}}{hypotenuse} \\ \Rightarrow\cos (45)=(y)/(24) \\ \Rightarrow y=24\cdot(\frac{1}{\sqrt[]{2}})=12\sqrt[]{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vfabupfsupsvk4hzpg110zyyg9f3hox5bt.png)
therefore, the values for x and y are:
![\begin{gathered} x=12\sqrt[]{2} \\ y=12\sqrt[]{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/lkmyigt5os4k6thzzff48vjezbqlsn8bp4.png)