Question

In an isosceles triangle with base ${}^{\prime }{a}^{\prime }$ and vertical angle ${20}^{\circ }$ and lateral side each ${}^{\prime }{b}^{\prime }$, show $a^3+b^3=3a{b}^2$.

Answer 1

If the vertical angle is $20$ then the other two are $80$
using cosine rule
$\cos 80=\frac{b^2+a^2+b^2}{2ab}$
$\sin 10=\frac{a}{2b}$
As $\sin 3x=3\sin x-4{\left(\sin x\right)}^3$
$\sin 30=\left(\frac{3a}{2b}\right)-\left(\frac{{4a}^3}{8b^3}\right)$
$\Rightarrow \frac{1}{2}=\frac{3a}{2b}-\frac{{4a}^3}{{8b}^2}\Rightarrow a^3+b^3=3a{b}^2$