Question

In triangle $\vartriangle ABC$, if $A+C=2B$ then $\dfrac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}$ is equals to
A. $2\cos \dfrac{A-C}{2}$
B. $\sin \dfrac{A+C}{2}$
C. $\sin \dfrac{A}{2}$
D. None of these.

Answer 1

Hint: We will use the angle sum property of the triangle and derive the equation and then substitute the given equation $A+C=2B$ in it and find the value of angle $B$. After this using cosine rule ${{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos B$ we will determine the length of the side $b$ by substituting the value of the angle $B$ derived.
Using the Law of sine, we will determine the value of the length of the sides $a,~b,~c$ and substitute in the equation of which the value we have to find.

Formula Used: $\sin{C}+\sin{D}=2\sin\dfrac{C+D}{2}\cos\dfrac{C-D}{2}$

Complete step by step solution: We are given a triangle $\vartriangle ABC$ and we have to determine the value of $\dfrac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}$ when $A+C=2B$.
We will use the angle sum property of the triangle according to which the sum of all the angles of a triangle is $\pi $. So,
$A+B+C=\pi $…. (i)
We will substitute $A+C=2B$ in equation (i).
$\begin{align}
  & 2B+B=\pi \\
 & 3B=\pi \\
 & B=\dfrac{\pi }{3} \\
\end{align}$
We will now use cosine rule,
$\begin{align}
  & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\cos \dfrac{\pi }{3} \\
 & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-2ac\times \dfrac{1}{2} \\
 & {{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac \\
 & b=\sqrt{{{a}^{2}}+{{c}^{2}}-ac}
\end{align}$
As we have to determine the value of $\dfrac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}$, and we have derived $b=\sqrt{{{a}^{2}}+{{c}^{2}}-ac}$, so we can write $\dfrac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}$as $\dfrac{a+c}{b}$.
So we will find the value of $\dfrac{a+c}{b}$ to determine the value of $\dfrac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}$.
We will now use sine rule or Law of sine,
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$
We can determine,
$a=k\sin A$……. (ii)
$c=k\sin C$……(iii)
$b=k\sin B$……. (iv)
Now we will substitute the value of equation (ii) ,(iii) and (iv) in $\dfrac{a+c}{b}$.
$\begin{align}
  & =\dfrac{k\sin A+k\sin C}{k\sin B} \\
 & =\dfrac{k(\sin A+\sin C)}{k\sin B} \\
 & =\dfrac{\sin A+\sin C}{\sin B}
\end{align}$
$=\dfrac{2\sin \dfrac{A+C}{2}\cos \dfrac{A-C}{2}}{\sin B}$
We will substitute $A+C=2B$ in the above equation.
$\begin{align}
  & =\dfrac{2\sin \dfrac{2B}{2}}{\sin B}.\cos \dfrac{A-C}{2} \\
 & =\dfrac{2\sin B}{\sin B}.\cos \dfrac{A-C}{2} \\
 & =2\cos \dfrac{A-C}{2} \\
\end{align}$

The value of $\dfrac{a+c}{\sqrt{{{a}^{2}}-ac+{{c}^{2}}}}$ when $A+C=2B$ in the triangle $\vartriangle ABC$ is $2\cos \dfrac{A-C}{2}$. Hence the correct option is (A).

Note: Law of sines is the ratio of the length of the side to the sine of its opposite angle and is the same for all of the three sides. The law of sines is written as $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ where $A$ is the angle opposite to side $a$, $B$ is the angle opposite to side $b$ and $C$ is the angle opposite to side $c$.