Question

In a triangle ABC if $\tan \left( \dfrac{A}{2} \right)=\dfrac{5}{6}$ and $\tan \left( \dfrac{B}{2} \right)=\dfrac{20}{37}$ , then sides a, b, c of the triangle are in?
(a) A.P.
(b) G.P.
(C) H.P.
(d) None of these

Answer 1

Hint:Use half angle formulae of tan function in terms of sides of triangle. It is given as
\[\begin{align}
  & \tan \dfrac{A}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{s\left( s-a \right)}} \\
 & \tan \dfrac{B}{2}=\sqrt{\dfrac{\left( s-c \right)\left( s-a \right)}{s\left( s-b \right)}} \\
 & \tan \dfrac{C}{2}=\sqrt{\dfrac{\left( s-a \right)\left( s-b \right)}{s\left( s-c \right)}} \\
\end{align}\]
where a, b, c are the sides of the triangle and the value of s is given as $\dfrac{a+b+c}{2}$ and A, B, C are the angles of the triangle.
From two equations with the given relations and solve them to get relations in (a, b, c).

Complete step-by-step answer:
Here we have a triangle ABC, with $\tan \left( \dfrac{A}{2} \right)=\dfrac{5}{6}$ and $\tan \left( \dfrac{B}{2} \right)=\dfrac{20}{37}$ , then we need to determine the relation among a, b and c

We know the half angle formulae for triangle for tan functions as:
\[\begin{align}
  & \tan \dfrac{A}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{s\left( s-a \right)}}......\left( i \right) \\
 & \tan \dfrac{B}{2}=\sqrt{\dfrac{\left( s-c \right)\left( s-a \right)}{s\left( s-b \right)}}......\left( ii \right) \\
\end{align}\]
Where the values of s = $\dfrac{a+b+c}{2}$ and a, b, c are the sides of the triangle, with opposite angles as A, B, C.
Now, as we know the values of $\tan \dfrac{A}{2}$ and $\tan \dfrac{B}{2}$ from the problem, so let us multiply equation (i) and (ii). So, we get
\[\tan \dfrac{A}{2}\tan \dfrac{B}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{s\left( s-a \right)}}\times \sqrt{\dfrac{\left( s-c \right)\left( s-a \right)}{s\left( s-b \right)}}\]
Now, we can cancel out (s – a) and (s – b) from numerator and denominator so we get
$\dfrac{5}{6}\times \dfrac{20}{37}=\sqrt{\dfrac{{{\left( s-c \right)}^{2}}}{{{s}^{2}}}}=\dfrac{s-c}{s}$
Now we know, the value of s is given as $\dfrac{a+b+c}{2}$. So, we get
$\begin{align}
  & \dfrac{50}{111}=\dfrac{\dfrac{a+b+c}{2}-c}{\left( \dfrac{a+b+c}{2} \right)} \\
 & \dfrac{50}{111}=\dfrac{\dfrac{\left( a+b+c-2c \right)}{2}}{\left( \dfrac{a+b+c}{2} \right)} \\
 & \dfrac{50}{111}=\dfrac{a+b-c}{a+b+c} \\
\end{align}$
Now, on cross multiplying the above equation, we get
50a + 50b + 50c = 111a + 111b – 111c
61a + 61b – 161c = 0………..(iii)
Similarly, let us divide the equation (i) and (ii) and put values of $\tan \dfrac{A}{2}$ and $\tan \dfrac{B}{2}$ .So, we get
\[\begin{align}
  & \dfrac{\tan \dfrac{A}{2}}{\tan \dfrac{B}{2}}=\dfrac{\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{s\left( s-a \right)}}}{\sqrt{\dfrac{\left( s-c \right)\left( s-a \right)}{s\left( s-b \right)}}} \\
 & \dfrac{\tan \dfrac{A}{2}}{\tan \dfrac{B}{2}}=\dfrac{\sqrt{\dfrac{s-b}{s-a}}}{\sqrt{\dfrac{s-a}{s-b}}}=\sqrt{\dfrac{s-b}{s-a}}\times \sqrt{\dfrac{s-b}{s-a}} \\
 & \dfrac{\dfrac{5}{6}}{\dfrac{20}{37}}=\sqrt{\dfrac{{{\left( s-b \right)}^{2}}}{{{\left( s-a \right)}^{2}}}} \\
 & \dfrac{5}{6}\times \dfrac{37}{20}=\dfrac{s-b}{s-a} \\
 & \dfrac{37}{24}=\dfrac{s-b}{s-a} \\
\end{align}\]
Now we can put value of s as \[\dfrac{a+b+c}{2}\] and hence, we get
$\begin{align}
  & \dfrac{37}{24}=\dfrac{\dfrac{a+b+c}{2}-b}{\dfrac{a+b+c}{2}-a}=\dfrac{\dfrac{a+b+c-2b}{2}}{\dfrac{a+b+c-2a}{2}} \\
 & \dfrac{37}{24}=\dfrac{a-b+c}{-a+b+c} \\
\end{align}$
On cross multiplying the above equation, we get
$\begin{align}
  & -37a+37b+37c=24a-24b+24c \\
 & -37a-24a+37b+24b+37c-24c=0 \\
 & -61a+61b+13c=0...........\left( iv \right) \\
\end{align}$
Now, on adding the equation (iii) and (iv), we get
\[\begin{align}
  & \underline{\begin{array}{*{35}{l}}
   {} & \text{ }61a & + & 61b & \text{ }- & 161c & = & 0 \\
   {} & -61a & + & 61b & \text{ }+ & 13c & = & 0 \\
   + & {} & {} & {} & {} & {} & {} & {} \\
\end{array}} \\
 & \begin{array}{*{35}{l}}
   {} & \text{ }0 & + & 122b & - & 148c & = & 0 \\
\end{array} \\
\end{align}\]
$b=\dfrac{148c}{122}=\dfrac{74c}{61}.......\left( v \right)$
On subtracting the equation (iii) and (iv), we get
\[\begin{align}
  & \underline{\begin{array}{*{35}{l}}
   {} & \text{ }61a & + & 61b & - & 161c & = & 0 \\
   {} & -61a & + & 61b & + & 13c & = & 0 \\
   - & {} & {} & {} & {} & {} & {} & {} \\
\end{array}} \\
 & \begin{array}{*{35}{l}}
   {} & 122a & + & \text{ }0b & - & 174c & = & 0 \\
\end{array} \\
\end{align}\]
$a=\dfrac{174c}{122}=\dfrac{87c}{61}......\left( vi \right)$
Now we can get value of $\dfrac{c}{61}$ from equation (v) and (vi) as
$\begin{align}
  & \dfrac{c}{61}=\dfrac{b}{74}=\dfrac{a}{87} \\
 & \Rightarrow \dfrac{a}{87}=\dfrac{b}{74}=\dfrac{c}{61} \\
\end{align}$
Now, let us suppose the value of all the above equal fractions be ‘k’. So, we get
$\dfrac{a}{87}=\dfrac{b}{74}=\dfrac{c}{61}=k$
Now, we get values of a, b and c in terms of k as
a = 87k, b = 74k, c = 61k
Now, we get that the terms
a, b, c i.e. 87k, 74k, 61k, has a common difference of ‘-7k’. $\left( 74k-87k=61k-74k=-7k \right)$ . It means a, b, c are in A.P.
So, option (a) is the correct answer.

Note: Another approach for the question would be that we can calculate $\cos A$ and $\cos B$ by using the formula
$\cos \theta =\dfrac{1-{{\tan }^{2}}\dfrac{\theta }{2}}{1+{{\tan }^{2}}\dfrac{\theta }{2}}$
And hence, apply the cosine formula with the given sides, cosine formula given as
$\cos \theta =\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$
Where $\theta $ is the angle between sides of length ‘b’ and ‘c’. So, get $\cos A$ and $\cos B$ and hence, use the above relations.
Direct applying identities with the chapter “properties of triangles” always makes the question flexible and less time taking, as done with this question in solution
Take care with the positions of terms involved with the identities mentioned in the solution. One may go wrong with these parts as well.