Question

Find csco, cose, and tan , where is the angle shown in the figure.Give exact values, not decimal approximations.

Answer 1

Let 'a' represent the unknown side of the triangle.

To solve for the unknown side, we will apply the Pythagoras theorem.

`a^2=b^2+c^2^{}`
`\begin{gathered} b=4 \\ c=5 \end{gathered}`
`\begin{gathered} a^2=4^2+5^2 \\ a=\sqrt[]{16+25} \\ a=\sqrt[]{41} \end{gathered}`

Let solve for cscθ

`\begin{gathered} \csc \theta=(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}`

Let us solve for cosθ

`\cos \theta=\frac{5}{\sqrt[]{41}}`
`\begin{gathered} \frac{5}{\sqrt[]{41}} \\ \text{Rationalising} \\ \frac{5}{\sqrt[]{41}}*\frac{\sqrt[]{41}}{\sqrt[]{41}}=\frac{5\sqrt[]{41}}{41} \end{gathered}`
`\cos \theta=\frac{5\sqrt[]{41}}{41}`

Let us solve for tanθ

`\tan \theta=(4)/(5)`