Question

In \(△ABC\), with usual notations, if \(\Delta\) denotes the area of triangle \(ABC\) then the value of \(2s(b+c-a)tan(\frac{A}{2})\) is equal to \(\ldots\)

• A. \(\Delta\)
• B. \(2\Delta\)
• C. \(3\Delta\)
• D. \(4\Delta\)

Hint:

Use tan(A/2) = Delta / (s(s-a)) and b+c-a = 2(s-a).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In a triangle with semi-perimeter \(s = \dfrac{a+b+c}{2}\), the half-angle formula for the tangent connects \(\tan\frac{A}{2}\) with the sides and the area \(\Delta\).

Step 2: Key Formula or Approach:
1. \(\tan\dfrac{A}{2} = \dfrac{\Delta}{s(s-a)}\).
2. \(b + c - a = 2s - 2a = 2(s-a)\).

Step 3: Detailed Explanation:
Substitute \(b + c - a = 2(s - a)\):
\[ 2s(b+c-a)\tan\frac{A}{2} = 2s \cdot 2(s-a) \cdot \tan\frac{A}{2} = 4s(s-a)\tan\frac{A}{2} \]
Now use the half-angle formula:
\[ 4s(s-a)\cdot\frac{\Delta}{s(s-a)} = 4\Delta \]
The factors \(s(s-a)\) cancel, leaving 4 times the area. Options (A), (B), (C) give \(\Delta\), \(2\Delta\) and \(3\Delta\), which would arise from missing the factor 2 in \(b+c-a\) or in the leading 2s.

Final Answer:
The value is \(4\Delta\), option (D). \[ \boxed{4\Delta \text{ (D)}} \]