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If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression $\dfrac{a}{c}\sin 2C + \dfrac{c}{a}\sin 2A$ is
A ) $\dfrac{1}{2}$
B ) $\dfrac{{\sqrt 3 }}{2}$
C ) $1$
D ) $\sqrt 3 $

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: Use the concept of the common difference of terms in an arithmetic progression and subsequently use the sine rule formula from properties of the triangle in order to convert the sides and angles in order to find out the value of the given expression.

Complete step by step solution:
We have been given that the angles A, B and C of a triangle are in an arithmetic progression.
Thus, using the common difference for terms in arithmetic progression, we get
$\begin{array}{l}
B - A = C - B\\
 \Rightarrow 2B = A + C\\
 \Rightarrow B = \dfrac{{A + C}}{2}.....\left( 1 \right)
\end{array}$
We also know that the sum of all three interior angles of a triangle is ${180^o}$
Therefore, $A + B + C = {180^o}$
Thus, we get
$A + C = {180^o} - B$
Using the above result and replacing the value of $A + C$ in equation $\left( 1 \right)$ , we get
$B = \dfrac{{{{180}^o} - B}}{2}$
Further simplifying, we get
$\begin{array}{l}
2B = {180^o} - B\\
 \Rightarrow 3B = {180^o}\\
 \Rightarrow B = {60^o}.....\left( 2 \right)
\end{array}$
Let us now consider the expression for which we need to find out the value
$\dfrac{a}{c}\sin 2C + \dfrac{c}{a}\sin 2A$
Using the formula of multiple angles $\sin 2\theta = 2\sin \theta \cos \theta $ , we get the above expression as
$\dfrac{a}{c}\left( {2\sin C\cos C} \right) + \dfrac{c}{a}\left( {2\sin A\cos A} \right)...\left( 3 \right)$
From the sine rule in properties of triangle, we know that
$\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} \Rightarrow \dfrac{a}{c} = \dfrac{{\sin A}}{{\sin C}}$
Using this ratio in the expression $\left( 3 \right)$ , we can rewrite the expression as
$\dfrac{{\sin A}}{{\sin C}}\left( {2\sin C\cos C} \right) + \dfrac{{\sin C}}{{\sin A}}\left( {2\sin A\cos A} \right)$
Cancelling out the terms, we get
$2\sin A\cos C + 2\sin C\cos A...\left( 4 \right)$
We know that $\sin \left( {x + y} \right) = \sin x\cos y + \sin y\cos x$
Using the above formula, we get the expression $\left( 4 \right)$ as
$2\sin \left( {A + C} \right)$
Substituting the value of $A + C$ from equation $\left( 1 \right)$ and then the value of $B = {60^o}$ in the above expression we get
$\begin{array}{l}
2\sin \left( {2B} \right)\\
 \Rightarrow 2\sin \left( {{{120}^o}} \right)\\
 \Rightarrow 2 \times \dfrac{{\sqrt 3 }}{2}\\
 \Rightarrow \sqrt 3
\end{array}$

Hence, the fourth option is correct.

Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant and is known as a common difference. The Law of Sines (or Sine Rule) is very useful for solving triangles. And it says that: When we divide side a by the sine of angle A it is equal to side b divided by the sine of angle B, and also equal to side c divided by the sine of angle C