Question

In a triangle $ABC$, if $\tan\dfrac{A}{2} = \dfrac{5}{6}$, and $\tan\dfrac{C}{2} = \dfrac{2}{5}$. Then find which of the following statements is true.
A. $a, c$, and $b$ are in AP.
B. $a, b$, and $c$ are in GP.
C. $b, a$, and $c$ are in AP.
D. $a, b$, and $c$ are in AP.

Answer 1

Hint: To calculate half angles, use the tangent of a trigonometric function and its half angle formula. Substituting the given values after multiplying the half angles To arrive at the required answer, simplify the equation in the end and check the common ratio or difference between $a, $b, and $c$.

Formula Used:
The half angle formula of tan for a triangle with sides $a, b$, and $c$, and the semi-perimeter $s$.
$\tan\dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} $
$\tan\dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{s\left( {s - b} \right)}}} $
$\tan\dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}}} $

Complete step by step solution:
Given: In a triangle $ABC$, $\tan\dfrac{A}{2} = \dfrac{5}{6}$, and $\tan\dfrac{C}{2} = \dfrac{2}{5}$.
Let $s$ be the semi-perimeter and $a, b$, and $c$ be the lengths of opposite sides of the angles $A,B$, and $C$ respectively of a triangle $ABC$.

Apply the half angle formula to calculate the values of $\tan\dfrac{A}{2}$, and $\tan\dfrac{C}{2}$.
$\tan\dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} $ $.....\left( 1 \right)$
$\tan\dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}}} $ $.....\left( 2 \right)$

Now multiply equation $\left( 1 \right)$ by equation $\left( 2 \right)$.
$\tan\dfrac{A}{2} \times \tan\dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}}} \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}}} $
Substitute the given values of the half angles.
$\dfrac{5}{6} \times \dfrac{2}{5} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{s\left( {s - a} \right)}} \times \dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}}} $ [Since $\sqrt a \times \sqrt b = \sqrt {ab} $]
Cancel out the common terms from numerator and denominator.
$\dfrac{1}{3} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - b} \right)}}{{s \times s}}} $
$ \Rightarrow $$\dfrac{1}{3} = \sqrt {\dfrac{{{{\left( {s - b} \right)}^2}}}{{{s^2}}}} $
$ \Rightarrow \dfrac{1}{3} = \dfrac{{\left( {s - b} \right)}}{s}$
Simplify the above equation.
$s = 3\left( {s - b} \right)$
$ \Rightarrow s = 3s - 3b$
$ \Rightarrow 2s = 3b$
Substitute the value of the semi-perimeter in the above equation.
$2\left( {\dfrac{{a + b + c}}{2}} \right) = 3b$
$ \Rightarrow a + b + c = 3b$
$ \Rightarrow a + c = 2b$
Since the sum of $a$ and $c$ is equal to the two times $b$.
This is the condition of the three numbers in arithmetic progression.
So, $a, b$, and $c$ are in arithmetic progression.

Option ‘D’ is correct

Note: An arithmetic series is a sequence in which each consecutive element is obtained by adding or subtracting the preceding element by a constant. The constant value is called a common difference.
The general form of an arithmetic series is $a, a + d, a + 2d, a + 3d,...$. Where $a$ is the first term and $d$ is a common difference.
If three numbers $x, y, z$ are in arithmetic progression, then $x + z = 2y$.