Question

If in a triangle \(ABC\), \(a=2\), \(b=3\) and \(c=4\), then
\[ \tan\left(\frac{A}{2}\right)= \]

• A. \(\sqrt{\dfrac{3}{15}}\)
• B. \(\sqrt{\dfrac{4}{15}}\)
• C. \(\sqrt{\dfrac{2}{15}}\)
• D. \(\sqrt{\dfrac{1}{15}}\)

Hint:

For triangle half-angle problems, remember the formula \(\displaystyle \tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\).

Answer 1

The Correct Option is D

Step 1: Use the half-angle formula in a triangle.
In any triangle,
\[ \tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \] where \(s\) is the semi-perimeter.

Step 2: Find the semi-perimeter.
Given,
\[ a=2,\quad b=3,\quad c=4 \]
Therefore,
\[ s=\frac{a+b+c}{2} \]
\[ =\frac{2+3+4}{2} \]
\[ =\frac92 \]

Step 3: Substitute in the formula.
Now,
\[ s-a=\frac92-2=\frac52 \] \[ s-b=\frac92-3=\frac32 \] \[ s-c=\frac92-4=\frac12 \]
Thus,
\[ \tan\frac{A}{2} = \sqrt{ \frac{\left(\frac32\right)\left(\frac12\right)} {\left(\frac92\right)\left(\frac52\right)} } \]
\[ = \sqrt{ \frac{\frac34}{\frac{45}{4}} } \]
\[ = \sqrt{\frac{3}{45}} \]
\[ = \sqrt{\frac{1}{15}} \]

Step 4: Final conclusion.
Hence,
\[ \boxed{\sqrt{\frac{1}{15}}} \]