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: Find the sides.
With $s = 36k$: $a = s - 11k = 25k$, $b = 36k - 12k = 24k$, $c = 36k - 13k = 23k$.

Step 2: Use cosine rule.
$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = \dfrac{576 + 529 - 625}{2 \cdot 24 \cdot 23} = \dfrac{480}{1104} = \dfrac{10}{23}$.

Step 3: Half angle.
$\tan^2\dfrac{A}{2} = \dfrac{1 - \cos A}{1 + \cos A} = \dfrac{13/23}{33/23} = \dfrac{13}{33}$.

Step 4: Find lambda.
$\lambda = 455 \times \dfrac{33}{13} = 1155$.

Final Answer:
Option (A). \[ \boxed{1155} \]