hello
to solve this problem, we would have to find the adjacent side of the bigger triangle using trigonometric ratios
using SOHCAHTOA
![\begin{gathered} \tan \theta=(opposite)/(adjacent) \\ \theta=58^0 \\ \text{opposite}=29 \\ \text{adjacent}=y \\ \tan 58=(29)/(y) \\ y=(29)/(\tan 58) \\ y=18.12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/egfpqrc9nj0wopjh27eo2p006yhdrx3wc5.png)
with the value of adjacent, we can as well assume it is the value of the opposite side of the smaller triangle.
to find the value of x, let's use the sine angle of this
![\begin{gathered} \sin \theta=(opposite)/(hypothenus) \\ \sin 29=(18.12)/(x) \\ x=(18.13)/(\sin 50) \\ x=23.6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zhiny1mxcnx8s6bxgna8i4hdqk7n4pak4o.png)
from the calculations above, the value of x is equal to 23.6 which corresponds to option B