Solving a triangle with the sine rule

Statement

In a triangle $\widehat{A}=35^\circ$, $\widehat{B}=80^\circ$ and the side opposite $\widehat{A}$ is $a=14\ \text{cm}$. Find the sides $b$ and $c$ and the area.

Solution

Third angle: $\widehat{C}=180^\circ-35^\circ-80^\circ=65^\circ$.

$b=\dfrac{a\sin\widehat{B}}{\sin\widehat{A}}=\dfrac{14\sin 80^\circ}{\sin 35^\circ}\approx \dfrac{14\cdot 0{.}98481}{0{.}57358}\approx 24{.}04$ cm.

$c=\dfrac{a\sin\widehat{C}}{\sin\widehat{A}}=\dfrac{14\sin 65^\circ}{\sin 35^\circ}\approx \dfrac{14\cdot 0{.}90631}{0{.}57358}\approx 22{.}12$ cm.

Area: $S=\dfrac12 a\,b\sin\widehat{C}=\dfrac12\cdot 14\cdot 24{.}04\cdot\sin 65^\circ\approx 152{.}5\ \text{cm}^2$.

$\boxed{b\approx 24{.}04\ \text{cm};\ c\approx 22{.}12\ \text{cm};\ S\approx 152{.}5\ \text{cm}^2}$