Question

In a triangle, if \[ b=5,\quad c=6,\quad \tan\frac{A}{2}=\frac{1}{\sqrt{2}}, \] then \[ a= \] is:

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

Hint:

When \(\tan\frac{A}{2}\) is given, first find \(\cos A\) using \[ \cos A=\frac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}} \] and then apply the cosine rule.

Answer 1

The Correct Option is A

Step 1: Use the half-angle relation.
Given, \[ \tan\frac{A}{2}=\frac{1}{\sqrt{2}} \] We know that \[ \cos A=\frac{1-\tan^2\frac{A}{2}}{1+\tan^2\frac{A}{2}} \] Now, \[ \tan^2\frac{A}{2}=\frac{1}{2} \] Therefore, \[ \cos A=\frac{1-\frac{1}{2}}{1+\frac{1}{2}} \] \[ \cos A=\frac{\frac{1}{2}}{\frac{3}{2}} \] \[ \cos A=\frac{1}{3} \]

Step 2: Apply cosine rule.
In a triangle, \[ a^2=b^2+c^2-2bc\cos A \] Substitute \[ b=5,\quad c=6,\quad \cos A=\frac{1}{3} \] \[ a^2=5^2+6^2-2(5)(6)\left(\frac{1}{3}\right) \] \[ a^2=25+36-20 \] \[ a^2=41 \]

Step 3: Find \(a\).
Since side length is positive, \[ a=\sqrt{41} \]

Step 4: Final conclusion.
Hence, \[ \boxed{\sqrt{41}} \]