Let 'a' represent the unknown side of the triangle.
To solve for the unknown side, we will apply the Pythagoras theorem.
[tex]a^2=b^2+c^2^{}[/tex][tex]\begin{gathered} b=4 \\ c=5 \end{gathered}[/tex][tex]\begin{gathered} a^2=4^2+5^2 \\ a=\sqrt[]{16+25} \\ a=\sqrt[]{41} \end{gathered}[/tex]
Let solve for cscθ
[tex]\begin{gathered} \csc \theta=\frac{1}{\sin \theta} \\ \sin \theta=\frac{4}{\sqrt[]{41}} \\ \text{Therefore,} \\ \csc \theta=\frac{1}{\frac{4}{\sqrt[]{41}}}=\frac{\sqrt[]{41}}{4} \end{gathered}[/tex]
Let us solve for cosθ
[tex]\cos \theta=\frac{5}{\sqrt[]{41}}[/tex][tex]\begin{gathered} \frac{5}{\sqrt[]{41}} \\ \text{Rationalising} \\ \frac{5}{\sqrt[]{41}}\times\frac{\sqrt[]{41}}{\sqrt[]{41}}=\frac{5\sqrt[]{41}}{41} \end{gathered}[/tex][tex]\cos \theta=\frac{5\sqrt[]{41}}{41}[/tex]
Let us solve for tanθ
[tex]\tan \theta=\frac{4}{5}[/tex]
