Oblicz pole

[LuckyLuck]

Oblicz pole trójkąta \(ABC\), w którym \(|AB|=7,\ |AC|=6\), \(\cos ( \angle ABC) = \frac{5}{7}\) .

[eresh]

Oblicz pole trójkąta Abc, w którym |AB|=7, |AC|=6, \(\cos ( \angle ABC) = \frac{5}{7}\) .

\(|AC|^2=|AB|^2+|BC|^2-|AB||BC|\cos\alpha\\
36=49+a^2-2\cdot 7\cdot a\cdot \frac{5}{7}\\
a^2-10a+13=0\\
a=5-2\sqrt{3}\;\; \vee \;\;\;a=5+2\sqrt{3}\)

\(\sin\alpha=\sqrt{1-(\frac{5}{7})^2}=\frac{2\sqrt{6}}{7}\)

dla \(a=5-2\sqrt{3}\)
\(P=\frac{1}{2}(5-2\sqrt{3})\cdot 7\cdot\frac{2\sqrt{6}}{7}=5\sqrt{6}-6\sqrt{2}\)

dla \(a=5+2\sqrt{3}\)
\(P=\frac{1}{2}(5+2\sqrt{3})\cdot 7\cdot\frac{2\sqrt{6}}{7}=5\sqrt{6}+6\sqrt{2}\)