Question

In a triangle ABC, $a = 2,\ b = 3,\ c = 4$, then $\tan\left(\dfrac{A}{2}\right) = $ ?

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

Hint:

Use $\tan\left(\dfrac{A}{2}\right) = \sqrt{ \dfrac{(s - b)(s - c)}{s(s - a)} }$ for quick computation in triangle problems.

Answer 1

The Correct Option is D

Use the half-angle formula: \[ \tan\left(\dfrac{A}{2}\right) = \sqrt{\dfrac{(s - b)(s - c)}{s(s - a)}} \] Where $s = \dfrac{a + b + c}{2} = \dfrac{2 + 3 + 4}{2} = \dfrac{9}{2}$ Now, \[ s - a = \dfrac{5}{2},\ s - b = \dfrac{3}{2},\ s - c = \dfrac{1}{2} \Rightarrow \tan\left(\dfrac{A}{2}\right) = \sqrt{ \dfrac{(3/2)(1/2)}{(9/2)(5/2)} } = \sqrt{ \dfrac{3}{90} } = \sqrt{ \dfrac{1}{15} } \]