Solving a Right Triangle

You can solve a right triangle with sides a, b, c and opposite angles α, β, γ, respectively, if you know the value of one side and one acute angle, or the value of any two sides.


\begin{example} To solve the right triangle with one side of length $c=2$\ and ... ...\frac{% 1}{9}\pi $\ $\left( =\,1.8794\right) $. \end{enumerate} \end{example}


\begin{example} To solve a right triangle given two sides, say $a=19$\ and $c=2... ... \frac{19}{23}$, $\beta =\arccos \frac{19}{23}.$ \end{enumerate} \end{example}

For a numerical result, evaluate these functions numerically, or choose Numeric rather than MenuDialogExact from the Solve submenu in the final step. You need to specify intervals for α and β, such as $\left[\vphantom{ \begin{array}{c} \sin \alpha =a/c \\ \alpha \in \left( 0,\pi /2\right) \end{array} }\right.$$\begin{array}{c} \sin \alpha =a/c \\ \alpha \in \left( 0,\pi /2\right) \end{array}$$\left.\vphantom{ \begin{array}{c} \sin \alpha =a/c \\ \alpha \in \left( 0,\pi /2\right) \end{array} }\right]$ and $\left[\vphantom{ \begin{array}{c} \cos \beta =a/c \\ \beta \in \left( 0,\pi /2\right) \end{array} }\right.$$\begin{array}{c} \cos \beta =a/c \\ \beta \in \left( 0,\pi /2\right) \end{array}$$\left.\vphantom{ \begin{array}{c} \cos \beta =a/c \\ \beta \in \left( 0,\pi /2\right) \end{array} }\right]$, before applying Solve + Numeric or you may get a solution greater than ${\frac{{\pi }}{{2}}}$. Specifying these intervals gives the solution

b =  12.96, α =  .9721 , β = .5987