Question

If in a triangle $ABC,a=5,b=4$ and $\cos (A-B)=\frac{31}{32}$. then the third side $c$ is equal to

• A. 3
• B. 6
• C. 7
• D. 9

Answer 1

The correct option is B 6

$\cos (A-B)=\frac{1-{\tan }^2\frac{A-B}{2}}{1+{\tan }^2\frac{A-B}{2}}\Rightarrow \frac{31}{32}=\frac{1-{\tan }^2\frac{A-B}{2}}{1+{\tan }^2\frac{A-B}{2}}\Rightarrow 63{\tan }^2\frac{A-B}{2}=1\Rightarrow \tan \frac{A-B}{2}=\frac{1}{\sqrt{63}}$
$\text{Now}\tan \frac{A-B}{2}=\frac{a-b}{a+b}\cot \frac{C}{2}\Rightarrow \frac{1}{\sqrt{63}}=\frac{5-4}{5+4}\cot \frac{C}{2}\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{63}}{9}\text{Also}\cos C=\frac{1-{\tan }^2\left(C/2\right)}{1+{\tan }^2\left(C/2\right)}$
$=\frac{1-\frac{63}{81}}{1+\frac{63}{81}}=\frac{18}{144}=\frac{1}{8}\therefore c^2=a^2+b^2-2ab\cos C=25+16-2\cdot 5\cdot 4\cdot \frac{1}{8}=36$
Hence $c=6$.