[tex]DC=30.430\approx30.4[/tex]
12) In this triangle, we need to find the projection DC. Let's start out finding the height, by using a trig ratio (sine):
[tex]\begin{gathered} \sin(54)=\frac{h}{20} \\ h=20\times\sin(54) \\ h=BD=16.18\approx16 \end{gathered}[/tex]
12.3) Based on another trig ratio we can write out the following formula:
[tex]\begin{gathered} \tan(28)=\frac{16.180}{DC} \\ DC=\frac{16.180}{\tan(28)} \\ DC=30.430 \end{gathered}[/tex]
