## Precalculus (6th Edition) Blitzer

The simplified value of $a$ is $8$ and the value of $b$ is $6$.
By using the Pythagoras identity in both triangles, we get \begin{align} & {{a}^{2}}+{{b}^{2}}={{10}^{2}} \\ & {{a}^{2}}+{{\left( b+9 \right)}^{2}}={{17}^{2}} \\ \end{align} Solve both of the above equations as shown below: ${{a}^{2}}+{{b}^{2}}=100$ (I) And \begin{align} & {{a}^{2}}+{{\left( b+9 \right)}^{2}}={{17}^{2}} \\ & {{a}^{2}}+{{b}^{2}}+81+18b=289 \end{align} Now, put the value of equation (I) and simplify this as shown below: \begin{align} & 100+81+18b=289 \\ & 18b=108 \\ & b=6 \end{align} Put the value of $b$ in equation (I) to get the value of $a$ as shown below: \begin{align} & {{a}^{2}}+{{6}^{2}}=100 \\ & {{a}^{2}}+36=100 \\ & {{a}^{2}}=64 \\ & a=8 \end{align} Hence, the value of $a$ is 8 and the value of $b$ is 6.