In triangle $ABC$, if $A$ is acute, $C$ is obtuse, $\sin A=\frac{3\sqrt{3}}{14},a=3$ and $b=5$, then $c=$

Given,

In triangle $ABC$, if $A$ is acute, $C$ is obtuse and $\sin A=\frac{3\sqrt{3}}{14}$,

Where $a=3$,$b=5$,

Now finding $\cos A=\sqrt{1-{\sin }^{2}A}=\sqrt{1-{\left(\frac{3\sqrt{3}}{14}\right)}^2}=\sqrt{\frac{169}{{14}^2}}=\frac{13}{14}$

Now using cosine formula we get,

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow \frac{13}{14}=\frac{5^2+c^2-3^2}{2\times 5c}$

$\Rightarrow \frac{13}{14}=\frac{16+c^2}{10c}$

$\Rightarrow 65c=7\left(16+c^2\right)$

$\Rightarrow 7c^2-65c+16\times 7=0$

$\Rightarrow 7c^2-49c-16c+16\times 7=0$

$\Rightarrow \left(7c-16\right)\left(c-7\right)=0$

$\Rightarrow c=\frac{16}{7} \text{or} c=7$

Since $C$ is obtuse so side $c$ will be greatest so $c=7$.