# What is the length of the shorter diagonal of the parallelogram if the lengths of the two adjacent sides of parallelogram ABCD are 8 and 14 units respectively and if the degree measure of the included angle is 60º?

Its typically helpful to start with a drawing. Here is $A B C D$ as given by the problem.
We are looking for the length of the shorter diagonal, which is segment $B D$. This segment forms a triangle with the two known sides. Since we know two sides and the angel connecting them, we can use the law of cosines to solve for the unknown segment.
The law of cosines tells us that ${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(C\right)$ for the triangle labeled above. If we choose our two known sides for $a$ and $b$ and our known angle for $C$, we can solve for the length of the diagonal, $c$.
${c}^{2} = {14}^{2} + {8}^{2} - 2 \left(14\right) \left(8\right) \cos \left({60}^{o}\right) \approx 12.17$