Let's begin by identifying key information given to us:
[tex]\begin{gathered} AB=20 \\ AC=x \\ BC=y \\ m\angle A=16^{\circ} \end{gathered}[/tex]
Since we have 1 known side & 1 known angle, we will solve using Trigonometric Ratio (SOHCAHTOA). In this case, we will use SOH
[tex]\begin{gathered} SOH\Rightarrow Sin\theta=\frac{opposite}{hypotenuse} \\ opposite=BC=y \\ hypotenuse=AB=20 \\ \theta=16^{\circ} \\ Sin16^{\circ}=\frac{y}{20} \\ y=Sin16^{\circ}(20)=5.51 \\ y=5.51 \end{gathered}[/tex]
