hello
the question given requests that we find a missing angle from a triangle
from the triangle above, we can simply use trigonometric ratios on this.
trigonometric ratios are SOHCAHTOA
[tex]\begin{gathered} \text{SOHCAHTOA} \\ \text{SOH}=\sin \theta=\frac{\text{opposite}}{\text{hypothenus}} \\ \text{CAH}=\cos \theta=\frac{\text{adjacent}}{\text{hypothenus}} \\ \text{TOA}=\tan \theta=\frac{\text{opposite}}{\text{adjacent}} \end{gathered}[/tex]
from the question given we have the value of opposite and adjacent and we can simply use tangent to find the missing angle
[tex]\begin{gathered} \tan \theta=\frac{opposite}{adjacent} \\ \text{opposite}=28 \\ \text{adjacent}=23 \\ \tan \theta=\frac{28}{23} \\ \tan \theta=1.21739 \\ \text{take the tan inverse of 1.217}39 \\ \theta=\tan ^{-1}1.21739 \\ \theta=50.6^0 \end{gathered}[/tex]
from the calculations above, the value of theta is equal to 50.6 degrees which corresponds to option B
