Question

In a triangle ABC , with usual notations, \(\tan \left(\frac{\mathrm{A}}{2}\right)=\frac{5}{6}, \tan \left(\frac{\mathrm{C}}{2}\right)=\frac{2}{5}\), then

• A. \(\mathrm{a},\mathrm{c},\mathrm{b}\) are in A.P.
• B. \(\mathrm{b},\mathrm{a},\mathrm{c}\) are in A.P.
• C. \(\mathrm{a},\mathrm{b},\mathrm{c}\) are in A.P.
• D. \(\mathrm{a},\mathrm{b},\mathrm{c}\) are in G.P.

Hint:

Use tan(A/2) tan(C/2) = (s - b)/s to relate the sides.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In a triangle, \(\tan\dfrac{A}{2}\tan\dfrac{C}{2}=\dfrac{s-b}{s}\), where \(s=\dfrac{a+b+c}{2}\).

Step 2: Calculation
\(\dfrac{5}{6}\times\dfrac{2}{5}=\dfrac{1}{3}=\dfrac{s-b}{s}\Rightarrow3s-3b=s\Rightarrow2s=3b\).

Step 3: Relation of Sides
\(a+b+c=3b\Rightarrow a+c=2b\).
So \(b\) is the mean of \(a\) and \(c\).

Step 4: Compare the options
The sides \(a,b,c\) are in A.P., which is the third option.

Final Answer:
The sides a, b, c are in arithmetic progression. \[ \boxed{\text{(C) }\text{a, b, c are in A.P.}} \]