Question

Please help me with this question from trig!!

Answer 1

`(\stackrel{adjacent}{4}~~,~~\stackrel{opposite}{5})\hspace{5em} \textit{let's find the \underline{hypotenuse}} \\\\\\ \begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2\implies c=√(a^2 + o^2) \end{array} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{4}\\ o=\stackrel{opposite}{5} \end{cases} \\\\\\ c=√( 4^2 + 5^2)\implies c=√( 16 + 25 ) \implies c=√( 41 ) \\\\[-0.35em] ~\dotfill`

`\sin(\alpha )=\cfrac{\stackrel{opposite}{5}}{\underset{hypotenuse}{√(41)}}\implies \sin(\alpha )=\cfrac{5√(41)}{41} \\\\\\ \cos(\alpha )=\cfrac{\stackrel{adjacent}{4}}{\underset{hypotenuse}{√(41)}}\implies \cos(\alpha )=\cfrac{4√(41)}{41} ~\hfill \tan(\alpha )=\cfrac{\stackrel{opposite}{5}}{\underset{adjacent}{4}}`

`\cot(\alpha )=\cfrac{\stackrel{adjacent}{4}}{\underset{opposite}{5}}~\hfill \sec(\alpha )=\cfrac{\stackrel{hypotenuse}{√(41)}}{\underset{adjacent}{4}}~\hfill \csc(\alpha )=\cfrac{\stackrel{hypotenuse}{√(41)}}{\underset{opposite}{5}}`