a fost răspuns

Sa se arate ca in orice triunghi ascutitunghic are relatia:
(b²+c²-a²)tgA=(a²+c²-b²)tgB

Răspuns :

Red12dog34ask
[tex]\left(b^2+c^2-a^2\right)\cdot\displaystyle\frac{\sin A}{\cos A}=\\=\left(b^2+c^2-a^2\right)\cdot\displaystyle\frac{\sin A}{\frac{b^2+c^2-a^2}{2bc}}=2bc\sin A=4S[/tex]

[tex]\left(a^2+c^2-b^2\right)\cdot\displaystyle\frac{\sin B}{\cos B}=\\=\left(a^2+c^2-b^2\right)\cdot\displaystyle\frac{\sin A}{\frac{a^2+c^2-b^2}{2ac}}=2ac\sin B=4S[/tex]

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