Question

Three sides of a triangle are $a,3a$ and $\sqrt{3}a$. Find the measurement of angle opposite to the largest side.

Answer 1

$\sin {}^{-1}(a-\frac{a^2}{3}+\frac{a^2}{9})+\cos (1+b+b^2)=\frac{\pi }{2}$

or, $\sin {}^{-1}\left(\frac{9a-3a^2+a^3}{9}\right)+\cos (1+b+b^2)=\frac{\pi }{2}$

or, $\cos (1+b+b^2)=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{9a-3a^2+a^3}{9}\right)$

Longest side is $3a$. Let opposite angle be $\alpha$.

$\cos \alpha =\frac{a^2+{\left(\sqrt{3}a\right)}^2-{\left(3a\right)}^2}{2\left(a\right)\left(\sqrt{3}a\right)}$

$=\frac{a^2+3a^2-9a^2}{2a^2\sqrt{3}}=-\frac{5a^2}{2a^2\sqrt{3}}=-\frac{5}{2\sqrt{3}}$

$\therefore \alpha ={\cos }^{-1}(-\frac{5}{2\sqrt{3}})$