When we have a right triangle like the one we have in the image, we can use the relationships between the legs and the hypotenuse.
The relationships are given by,
![\sin 45=(x)/(y),\text{ }\cos 45=\frac{\frac{\sqrt[]{2}}{2}}{y}](https://img.qammunity.org/2023/formulas/mathematics/college/5lu89nbkcr5kyjq8vux7e2aqqpmid64eij.png)
We need to find x, however, to find x we must find y first,
We solve for y,
![\begin{gathered} y=\frac{\frac{\sqrt[]{2}}{2}}{\cos 45} \\ y=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tspgxpd3ombf4zmiyuuqe42kavt77pf2i5.png)
Now that we have the value of y, we can find x,
![\begin{gathered} x=\sin 45* y \\ x=\sin 45*1 \\ x=\frac{\sqrt[]{2}}{2} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hifp284223wyc9h4n0efh8lgrryhb4gm8m.png)