Question

In triangle $ABC,a=5,b=4$ and $\cos (A+B)=\frac{31}{32}$. In this triangle, $c=$

• A. $\sqrt{6}$
• B. $36$
• C. $6$
• D. $\frac{\sqrt{319}}{2}$

Answer 1

The correct option is D $\frac{\sqrt{319}}{2}$

In $△ABC$-

$A+B+C=\pi$

$A+B=\pi -C$

$\cos (A+B)=-\cos C=\frac{31}{32}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{-31}{32}$

$\frac{25+16-c^2}{\mathrm{2.5.4}}=\frac{-31}{32}$

$\frac{41-c^2}{8.5}=\frac{-31}{32}$

$164-4c^2=-155$

$4c^2=319$

$c=\frac{\sqrt{319}}{2}$