In an isosceles triangle the vertex angle is equal to c and the area is S. Find the length of the base of the triangle.

Hint: Isosceles triangle has two equal sides and altitude from the vertex of the equal sides to the base will divide the triangle in two congruent triangles. Area of a triangle is given as
$=\dfrac{1}{2}\times base\times height$
Use the relation $\tan \theta =\dfrac{Perpendicular}{Base}$ and use the conditions given in the triangle to get the answer.

Complete step-by-step answer:
Let us suppose the isosceles triangle is given as $\Delta ABC$ as given below

where, AB = BC and area of triangle ABC is ‘s’ as given in the problem and vertex angle i.e. angle A, is given as ‘c’ and hence we need to calculate the base of triangle i.e. length BC.
As, we know area of triangle is given as
Area of triangle = $\dfrac{1}{2}\times base\times height........\left( i \right)$
So, let us draw a perpendicular from A to side BC. Hence, we get diagram as



Now, as we know ABC is representing a isosceles triangle with equal sides as AB and AC and AM is acting as a perpendicular from $\Delta ABM$ and $\Delta AMC$ as
AB = AC (Given)
$\angle AMB=\angle AMC$ (Construction)
AM = AM (Common)
Hence, $\Delta ABM\cong \Delta ACM$ by RHS criteria. So, $\angle BAM=\angle CAM$ by C.P.C.T rule.
So, AM will divide the vertex angle i.e. ‘c’ in two parts. So, we get
$\angle BAM=\angle CAM=\dfrac{c}{2}.........\left( ii \right)$
And BM = MC by C.P.C.T as well. So, we get that M is a perpendicular bisector of side BC. So, we get
$BM=MC=\dfrac{BC}{2}.........\left( iii \right)$
Now, take tan of angle $\dfrac{c}{2}$ in triangle ABM as
$\tan \left( \dfrac{c}{2} \right)=\dfrac{Perpendicular}{Base}=\dfrac{BM}{AM}$
We know, $BM=\dfrac{BC}{2}.$ We get
\[\begin{align}
  & \tan \left( \dfrac{c}{2} \right)=\dfrac{BC}{2AM} \\
 & \Rightarrow AM=\dfrac{BC}{2\tan \left( \dfrac{c}{2} \right)}........\left( iv \right) \\
\end{align}\]
We know area of triangle ABC is given as
Area of $\Delta ABC=\dfrac{1}{2}\times AM\times BC$
We know the area of $\Delta ABC$ is given as ‘s’ in the problem and we can replace AM by $\dfrac{BC}{2\tan \left( \dfrac{c}{2} \right)}$ from the equation (iv). So, we get
$\begin{align}
  & s=\dfrac{1}{2}\times \dfrac{BC}{2\tan \left( \dfrac{C}{2} \right)}\times BC \\
 & 4s\tan \left( \dfrac{c}{2} \right)={{\left( BC \right)}^{2}} \\
 & \Rightarrow {{\left( BC \right)}^{2}}=4s\tan \left( \dfrac{c}{2} \right) \\
\end{align}$
Taking root on both sides, we get
$BC=\sqrt{4s\tan \dfrac{C}{2}}=2\sqrt{s\tan \dfrac{c}{2}}$ .
Hence, the length of the base is given as $2\sqrt{s\tan \dfrac{c}{2}}$ . So, \[2\sqrt{s\tan \dfrac{c}{2}}\] is the answer.

Note: One may use Pythagoras theorem to get the base but need to use formula of area of a triangle as $\dfrac{1}{2}ab\sin \theta $ , where $\theta $ is the angle between side ‘a’ and ‘b’. It means apply Pythagora's theorem in $\Delta ABM$ to get AB in terms of BM and hence, apply the above-mentioned formula to get area.