## Precalculus (6th Edition) Blitzer

The length of the sides of triangle are $a=11.6$ and $b=23.9$
Let the lower left point and lower right point of given triangle be $D\text{ and }E$, respectively. Now by applying law of cosines in triangle $ADE$, we get \begin{align} & \cos A=\frac{{{17}^{2}}+{{26}^{2}}-{{37.5}^{2}}}{2\cdot 17\cdot 26} \\ & \cos A=\frac{289+676-1406.25}{884} \\ & A={{\cos }^{-1}}\left( -.499 \right) \\ & A\simeq 120{}^\circ \end{align} Similarly, we will apply the law of cosines in triangle $ABE$ \begin{align} & \cos A=\frac{{{26}^{2}}+{{21.5}^{2}}-{{13.5}^{2}}}{2\cdot 26\cdot 21\cdot 5} \\ & \cos A=\frac{676+462.25-182.25}{1118} \\ & A={{\cos }^{-1}}.855 \\ & A\simeq 31{}^\circ \end{align} After that we will apply the sine law \begin{align} & \frac{13.5}{\sin 31{}^\circ }=\frac{21.5}{\sin E} \\ & E=56{}^\circ \\ \end{align} And \begin{align} & B=180{}^\circ -56{}^\circ -31{}^\circ \\ & B=93{}^\circ \\ \end{align} Now, consider the triangle $ACB$, to find out the angle $\angle CAB$ \begin{align} & \angle CAB=180-\angle BAE-\angle EAD \\ & \angle CAB=180{}^\circ -120{}^\circ -31{}^\circ \\ & \angle CAB=29{}^\circ \\ \end{align} Consider the triangle $ACB$, to find out the angle $\angle CBA$ \begin{align} & \angle CBA=180-\angle EBA \\ & \angle CAB=180{}^\circ -93{}^\circ \\ & \angle CAB=87{}^\circ \\ \end{align} And \begin{align} & C=180{}^\circ -87{}^\circ -29{}^\circ \\ & C=64{}^\circ \\ \end{align} Now by using sine law we get, \begin{align} & \frac{a}{\sin 29{}^\circ }=\frac{21.5}{\sin 64{}^\circ } \\ & a\simeq 11.6 \end{align} Again, by applying sine law we will find out the side of the triangle. \begin{align} & \frac{b}{\sin 87{}^\circ }=\frac{21.5}{\sin 64{}^\circ } \\ & b\simeq 23.9 \end{align}