Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$. What is the length of the shorter diagonal of $ABCD$? $\textbf{(A) }\frac{31}7 \qquad$ $ \textbf{(B) }\frac{33}7 \qquad$ $ \textbf{(C) }5 \qquad$ $ \textbf{(D) }\frac{39}7 \qquad$ $ \textbf{(E) }\frac{41}7 \qquad$
Solution
In $\triangle CDA$, $AC^2=3^2+5^2-2\cdot3\cdot5\cos120^\circ=9+25+15=49$, so $AC=7$.
$ABCD$ is cyclic, so $\angle ABC=180^\circ-120^\circ=60^\circ$. In $\triangle ABC$: $7^2=AB^2+3^2-2\cdot AB\cdot3\cos60^\circ$, i.e. $AB^2-3AB-40=0$, so $AB=8$.
By Ptolemy's theorem, $AC\cdot BD=AB\cdot CD+BC\cdot DA=8\cdot3+3\cdot5=39$, so $BD=\dfrac{39}7\approx5.6<7$.
