## Precalculus (6th Edition) Blitzer

${{\operatorname{sinB}}^{\circ }}>1$ Thus, no real value for B exists.
\begin{align} & \frac{51}{\sin {{75}^{\circ }}}=\frac{71}{{{\operatorname{sinB}}^{\circ }}} \\ & 51{{\operatorname{sinB}}^{\circ }}=71\sin {{75}^{\circ }} \\ & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\sin {{75}^{\circ }} \\ & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\sin \left( 45+30 \right) \end{align} Use the identity $\sin \left( \alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$. \begin{align} & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\left[ \sin 45\cdot cos30+cos45\cdot sin30 \right] \\ & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\left[ \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot \frac{1}{2} \right] \\ & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\left[ \frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{2\sqrt{2}} \right] \\ & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\left[ \frac{2\sqrt{3}+1}{2\sqrt{2}} \right] \\ \end{align} Simplified, \begin{align} & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\left[ \frac{2\left( 1.732 \right)+1}{2\left( 1.4142 \right)} \right] \\ & {{\operatorname{sinB}}^{\circ }}=\frac{71}{51}\times \frac{4.464}{2.284} \\ & {{\operatorname{sinB}}^{\circ }}=\frac{316.944}{116.484} \\ & {{\operatorname{sinB}}^{\circ }}=1.344 \\ \end{align} We see that ${{\operatorname{sinB}}^{\circ }}>1$. So, no real value of $B$ exists. Thus, ${{\operatorname{sinB}}^{\circ }}>1$ No real value for B exists.