Solving a triangle with the law of sines

Text

A triangle has angles $\widehat A=35^\circ$, $\widehat B=65^\circ$ and side $a=10$ opposite $\widehat A$. Find the other two sides $b$ and $c$.

Solution

The third angle is $\widehat C=180^\circ-35^\circ-65^\circ=80^\circ$. By the law of sines $\dfrac{a}{\sin\widehat A}=\dfrac{b}{\sin\widehat B}=\dfrac{c}{\sin\widehat C}$: $b=a\,\frac{\sin\widehat B}{\sin\widehat A}=10\cdot\frac{\sin 65^\circ}{\sin 35^\circ}\approx 10\cdot\frac{0.9063}{0.5736}\approx 15.80,$ $c=a\,\frac{\sin\widehat C}{\sin\widehat A}=10\cdot\frac{\sin 80^\circ}{\sin 35^\circ}\approx 10\cdot\frac{0.9848}{0.5736}\approx 17.17.$ $\boxed{\,b\approx 15.80,\quad c\approx 17.17\,}$