## Precalculus (6th Edition) Blitzer

Angles $23^\circ,67^\circ,90^\circ$ and sides $12, 28.27, 30.71$
Step 1. With $C=90^\circ, A=23^\circ$, we have $B=90^\circ-23^\circ=67^\circ$ Step 2. In the right triangle, we have $sinA=\frac{a}{c}$; thus $sin23^\circ=\frac{12}{c}$ which gives $c=\frac{12}{sin23^\circ}\approx30.71$ Step 3. In the right triangle, we have $tanA=\frac{a}{b}$; thus $tan23^\circ=\frac{12}{b}$, which gives $b=\frac{12}{tan23^\circ}\approx28.27$ Step 4. The angles of the triangle are $23^\circ,67^\circ,90^\circ$ and the sides are $12, 28.27, 30.71$