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: Test with an equilateral triangle:
Let $a=b=c=2$. Then $s = 3$, $A = 60^\circ$, $\tan 30^\circ = \frac{1}{\sqrt3}$, and $\Delta = \frac{\sqrt3}{4}\cdot 4 = \sqrt3$.

Step 2: Evaluate the expression:
$2s(b+c-a)\tan\frac{A}{2} = 2\cdot3\cdot(2+2-2)\cdot\frac{1}{\sqrt3} = \frac{12}{\sqrt3} = 4\sqrt3$.

Step 3: Compare:
$4\sqrt3 = 4\Delta$ since $\Delta = \sqrt3$. This agrees with the general result obtained using $\tan\frac A2 = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.

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