Question

In a triangle ABC, with usual notations if \(\frac{s-a}{11} = \frac{s-b}{12} = \frac{s-c}{13}\) and \(λtan^2\frac{A}{2} = 455\), then \(λ =\)

• A. \(1155\)
• B. \(1255\)
• C. \(1355\)
• D. \(1055\)

Hint:

Let s-a, s-b, s-c be proportional to 11, 12, 13 and use the half-angle formula for tan(A/2).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In triangle ABC, \(s\) is the semi-perimeter. The half-angle formula is \(\tan^2\dfrac{A}{2} = \dfrac{(s-b)(s-c)}{s(s-a)}\).

Step 2: Find the ratios.
Let \(s - a = 11k\), \(s - b = 12k\), \(s - c = 13k\). Adding gives \(3s - 2s = s = 36k\).

Step 3: Apply the formula.
\[ \tan^2\frac{A}{2} = \frac{12k \times 13k}{36k \times 11k} = \frac{156}{396} = \frac{13}{33} \]

Step 4: Solve for \(\lambda\).
\[ \lambda\cdot\frac{13}{33} = 455 \Rightarrow \lambda = \frac{455 \times 33}{13} = 35 \times 33 = 1155 \]

Final Answer:
\(\lambda = 1155\), option (A). \[ \boxed{1155} \]