## Precalculus (6th Edition) Blitzer

$b\approx7.6, A\approx67^\circ, C\approx23^\circ$
Step 1. Based on the given conditions, using the Pythagorean Theorem, we have $b=\sqrt {a^2+c^2}=\sqrt {7^2+3^2}=\sqrt {58}\approx7.6$ Step 2. For angle $A$, in the right triangle, we have $tanA=\frac{a}{c}=\frac{7}{3}$, thus $A=tan^{-1}(\frac{7}{3})\approx67^\circ$ Step 3. We can find the angle as $C=90^\circ-67^\circ=23^\circ$ Step 4. We solved the triangle with $b\approx7.6, A\approx67^\circ, C\approx23^\circ$