Na figura, um triângulo genérico $\,\triangle ABC\,$ onde deseja-se a medida do ângulo $\,\hat{A}\,$.
De acordo com a lei dos cossenos temos:
$\;a^2\,=\,b^2\,+\,c^2\,-\,2bc\centerdot (cos\hat{A})\;(I)$
Mas (conforme o enunciado), $\,a\,=\,\dfrac{7c}{3}\,$ e $\,b\,=\,\dfrac{8c}{3}\,$, substituindo em (I)
$\,\left( \dfrac{7c}{3}\right)^{\large 2}\;=\;\left( \dfrac{8c}{3} \right)^{\large 2}\,+\,c^{\large 2}\, -\,2\centerdot \left( \dfrac{8c}{3} \right)\centerdot c \centerdot cos\hat{A}\;\Rightarrow\,$
$\,\Rightarrow\,\left( \dfrac{49c^{\large 2}}{9}\right)\;=\;\left( \dfrac{64c^{\large 2}}{9} \right)^\,+\,\dfrac{9c^{\large 2}}{9}\, -\,2\centerdot \left( \dfrac{24c^{\large 2}}{9} \right)\centerdot cos\hat{A}\,\Rightarrow\,$
$\,\Rightarrow\,49\left( \dfrac{c^{\large 2}}{9}\right)\;=\;64\left( \dfrac{c^{\large 2}}{9} \right)\,+\,9\left(\dfrac{c^{\large 2}}{9}\right)\, -\,2\centerdot 24 \centerdot cos\hat{A}\left( \dfrac{c^{\large 2}}{9} \right)\,$
● dividir a igualdade por c2/9
$\,\Rightarrow\,49\;=\;64\,+\,9\, -\,2\centerdot 24 \centerdot cos\hat{A}\,$
$\,\Rightarrow\,-cos\hat{A}\,=\,\dfrac{49\,-\,64\,-\,9}{2\centerdot 24}\,\Rightarrow\,$
$\,\Rightarrow\,cos\hat{A}\,=\,\dfrac{24}{48}\,\Rightarrow\,cos\hat{A}\,=\,\dfrac{1}{2}\;\Rightarrow\; \hat{A}\,=\,60^o$
