Aus Online Mathematik Brückenkurs 1

An often-used technique to calculate \displaystyle \text{cos }v and \displaystyle \text{tan }v, given the sine value of an acute angle, is to draw the angle \displaystyle v in a right-angled triangle which has two sides arranged so that \displaystyle \text{sin }v={5}/{7}\;.

Using Pythagoras' theorem, we can determine the length of the third side in the triangle.

\displaystyle x^{2}+5^{2}=7^{2} which gives that \displaystyle x=\sqrt{7^{2}-5^{2}}=\sqrt{24}=2\sqrt{6}

Then, using the definition of cosine and tangent,

\displaystyle \begin{align} & \cos v=\frac{x}{7}=\frac{2\sqrt{6}}{7}, \\ & \tan v=\frac{5}{x}=\frac{5}{2\sqrt{6}} \\ \end{align}

NOTE: Note that the right-angled triangle that we use is just a tool and has nothing to do with the triangle that is referred to in the question.