[tex3]\mathsf{
L. cossenos \triangle BPC\\
BC^2=25k^2+49k^2−2⋅5k⋅7k⋅cos(x)\implies\\
BC^2=74k^2−70k^2⋅cos(x)(I)\\
L. cossenos \triangle BPA\\
AB^2=25k^2+k^2−2⋅5k⋅k⋅cos(π−x)\\
AB^2=26k^2+10k^2⋅cos(x)(II)
(I)+(II): \\
BC^2+AB^2=26k^2+10k^2⋅cos(x)+74k^2−70k^2⋅cos(x)\\
AC^2=100k^2−60k^2⋅cos(x)\\
64k^2=100k^2−60k^2⋅cos(x)\\
\frac{36}{60}=cos(x)\implies co(x) = \frac{3}{5} \therefore \boxed{\color{red}{x = 53^o}}}[/tex3]
(Solução: Ittalo25 -
viewtopic.php?f=4&t=56017&p=146787&hili ... na#p146787)