If in a triangle $ABC$ right angled at $B$, $s-a=3,\,s-c=2$, then the values of $a$ and $c$ are respectively.
A. $2,3$
B. $3,4$
C. $4,3$
D. $6,8$

Hint: To solve this question, we will use the formula of semi perimeter of the triangle. We will substitute the value of $s$ in given equations $s-a=3,\,s-c=2$ and simplify it. After simplifying we will get two equations from which we will get the value of one of the variables. Then using substitution method and simplifying we will get the value of $a$ and $c$

Formula Used: Semi perimeter $s=\dfrac{a+b+c}{2}$.

Complete step by step solution: We are given a triangle $ABC$ right angled at $B$ with $s-a=3,\,s-c=2$ and we have to determine the value of $a$ and $c$.
We will first take the equation $s-a=3$ and substitute the value of $s$.
$\begin{align}
  & \dfrac{a+b+c}{2}-a=3 \\
 & a+b+c-2a=6 \\
 & b+c-a=6.....(i)
\end{align}$
Now we will take the equation $s-c=2$ and substitute the value of $s$.
$\begin{align}
  & s-c=2 \\
 & \dfrac{a+b+c}{2}-c=2 \\
 & a+b-c=4...(ii)
\end{align}$
Adding equations (i) and (ii).
$\begin{align}
  & b+c-a+a+b-c=6+4 \\
 & 2b=10 \\
 & b=5
\end{align}$
As a given triangle$ABC$ right angled at $B$, we can use the Pythagoras theorem. So,
$\begin{align}
  & {{5}^{2}}={{a}^{2}}+{{c}^{2}} \\
 & 25={{a}^{2}}+{{c}^{2}}....(iii) \\
\end{align}$
Using equation (ii),
$\begin{align}
  & a-c+5=4 \\
 & a-c=-1.....(iv) \\
\end{align}$
Squaring both sides,
$\begin{align}
  & {{(a-c)}^{2}}=1 \\
 & {{a}^{2}}+{{c}^{2}}-2ac=1 \\
\end{align}$
We will now substitute the equation (iii) in the above equation.
$\begin{align}
  & 25-2ac=1 \\
 & 24=2ac \\
 & ac=12 \\
 & a=\dfrac{12}{c}
\end{align}$
We will now substitute the value $a=\dfrac{12}{c}$ in equation (iv).
$\begin{align}
  & \dfrac{12}{c}-c=-1 \\
 & 12-{{c}^{2}}=-c \\
 & {{c}^{2}}-c-12=0 \\
 & (c-4)(c+3)=0 \\
 & c=-3,4
\end{align}$
We will consider the value of $c$ as $c=4$ because the value of the length cannot be negative.
Now we will substitute the value of $c$ in equation (iv) and derive the value of $a$.
$\begin{align}
  & a-c=-1 \\
 & a-4=-1 \\
 & a=3
\end{align}$

The value of $a$ and $c$ are $a=3$ and $c=4$ when the triangle $ABC$ is right angled at $B$ and $s-a=3,\,s-c=2$. Hence the correct option is (B).

Note: The Pythagoras theorem is only applicable for a right angled triangle and it depicts the relationship between the three sides of the triangle.