Question

In a triangle, if \( b = 5 \), \( c = 6 \), and \( \tan\left(\frac{A}{2}\right) = \frac{1}{\sqrt{2}} \), then the side \( a = \) ?

• A. \( \sqrt{41} \)
• B. \( \sqrt{21} \)
• C. \( \sqrt{14} \)
• D. \( \sqrt{22} \)

Hint:

Use the tangent-half angle formula in triangle side problems. Substitute using semi-perimeter and simplify algebraically.

Answer 1

The Correct Option is A

Use the identity: \[ \tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} \] Let \( a \) be unknown, and compute \( s = \frac{a + 5 + 6}{2} = \frac{a + 11}{2} \) Now plug into the formula: \[ \frac{1}{\sqrt{2}} = \sqrt{ \frac{(s - 5)(s - 6)}{s(s - a)} } \] Square both sides: \[ \frac{1}{2} = \frac{(s - 5)(s - 6)}{s(s - a)} \] Now plug \( s = \frac{a + 11}{2} \) into all terms and simplify: Let’s just substitute and simplify: \[ s = \frac{a + 11}{2}, \quad s - 5 = \frac{a + 1}{2}, \quad s - 6 = \frac{a - 1}{2}, \quad s - a = \frac{11 - a}{2} \] So: \[ \begin{align} \frac{1}{2} = \frac{\left(\frac{a + 1}{2}\right)\left(\frac{a - 1}{2}\right)}{\left(\frac{a + 11}{2}\right)\left(\frac{11 - a}{2}\right)} = \frac{(a^2 - 1)/4}{(121 - a^2)/4} = \frac{a^2 - 1}{121 - a^2} \] Cross-multiply: \[ a^2 - 1 = \frac{1}{2}(121 - a^2) \Rightarrow 2(a^2 - 1) = 121 - a^2 \Rightarrow 2a^2 - 2 = 121 - a^2 \Rightarrow 3a^2 = 123 \Rightarrow a^2 = 41 \Rightarrow a = \sqrt{41} \]