Question

If in a $△ABC$, $a=5$, $b=4$ and $\cos (A-B)=\frac{31}{32}$, then

• A. $c=6$
• B. $\sin A=\left(\frac{5\sqrt{7}}{16}\right)$
• C. area of $△ABC=\frac{15\sqrt{7}}{4}$
• D. $c=8$

Answer 1

The correct option is A $c=6$

Using $cos\left(\frac{A-B}{2}\right)=\frac{1-Ta{n}^{2}A-B/2}{1+Ta{n}^{2}A-B/2}=\frac{31}{32}$

$\Rightarrow Tan\frac{A-B}{2}=\frac{1}{\sqrt{63}}$

Now,

$Tan\frac{A-B}{2}=\frac{a-b}{a+b}cot\frac{c}{2}\Rightarrow Tan\frac{c}{2}=\frac{\sqrt{63}}{9}$

Now,

$cosc=\frac{1-Ta{n}^{2}C/2}{1+Ta{n}^{2}C/2}\Rightarrow cosc=\frac{1}{8}$

Now,

Using $c^2+a^2+b^2-2abcosc=36$

$c=6$