Question

If in a triangle $a=5,b=4$, and $cos(A-B)=\frac{31}{32}$, prove that the third side c will be 6.

Answer 1

$tan\frac{\alpha }{2}=(\frac{1-cos\alpha }{1+cos\alpha }{)}^{1/2}=(\frac{1-\left(31/32\right)}{1+\left(31/32\right)}{)}^{1/2}$

$=\frac{1}{\sqrt{63}}=\frac{1}{3\sqrt{7}}$ where $\alpha =A-B$

Now $tan\frac{A-B}{2}=\frac{a-b}{a+b}cos\frac{C}{2}$ ( Nap. Analogy)

put $a=5,b=4$ $\therefore \frac{1}{3\sqrt{7}}=\frac{1}{9}cot\frac{C}{2}$ or $tan\frac{C}{2}=\frac{\sqrt{7}}{3}$

Now $cosC=\frac{1-ta{n}^2\left(C/2\right)}{1+ta{n}^2\left(C/2\right)}=\frac{1-7/9}{1+7/9}=\frac{1}{8}$

But $c^2=a^2+b^2-2abcosC=36\therefore c=6$

Hence proved.