Question

In a triangle ABC if $tan\frac{1}{2}A=\frac{5}{6}$ and $tan\frac{1}{2}B=\frac{20}{37}$ find $tan\frac{1}{2}C,$ and prove that in this triangle $a+c=2b.$

Answer 1

We have$tan\frac{C}{2}=tan({90}^0-\frac{A+B}{2})=cot(\frac{A}{2}+\frac{B}{2})$

$=\frac{cot\frac{A}{2}cot\frac{B}{2}-1}{cot\frac{B}{2}+cot\frac{A}{2}}$

$=\frac{\frac{6}{5}.\frac{37}{20}-1}{\frac{37}{20}+\frac{6}{5}}$

=$\frac{222-100}{5\times 61}=\frac{122}{5\times 61}=\frac{2\times 61}{5\times 61}=\frac{2}{5}$.

Again $tan\frac{C}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

Hence $\frac{5}{6}.\frac{2}{5}=\frac{s-b}{s}$

$\therefore 3s-3b=sora+c=2b$.