Side and area with the law of cosines

Text

In a triangle two sides measure $a=7$ and $b=9$ and the included angle is $\widehat C=40^\circ$. Find the third side $c$ and the area of the triangle.

Solution

Law of cosines for the side opposite $\widehat C$: $c^{2}=a^{2}+b^{2}-2ab\cos\widehat C=49+81-2\cdot 7\cdot 9\cos 40^\circ\approx 130-96.52=33.48,$ so $c\approx\sqrt{33.48}\approx 5.79$. Area from two sides and the included angle: $A=\tfrac12\,ab\sin\widehat C=\tfrac12\cdot 7\cdot 9\cdot\sin 40^\circ\approx 31.5\cdot 0.6428\approx 20.25.$ $\boxed{\,c\approx 5.79,\quad A\approx 20.25\,}$