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$\Delta ABC,$ biết nó thỏa các điều kiện sau:

$\left\{ \begin{array}{l}\widehat{A}>\widehat{B}>\widehat{C}\\\cos3A+\cos3B+\cos3C=1\\\sin5A+\sin5B+\sin5C=0\end{array} \right.$

khangnguyenthanh · Answer 1 · 10/18/2013 6:05:51 PM

$1=\cos 3A+\cos 3B+\cos 3C=2\cos \frac{3(A+B)}{2}\cos \frac{3(A-B)}{2}+1-2\sin^2 \frac{3C}{2}$

$\Rightarrow 1=-2\sin \frac{3C}{2}\cos \frac{3(A-B)}{2}=1-2\sin \frac{3C}{2}\left(\cos \frac{3(A+B)}{2}-\cos \frac{3(A-B)}{2} \right)$

$\Rightarrow 1=1+4\sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}\Rightarrow \sin \frac{3A}{2}\sin \frac{3B}{2}\sin \frac{3C}{2}=0$

$\Rightarrow \left[\begin{array}{l}\angle A=120^o\\\angle B=120^o\\\angle C=120^o\end{array}\right. \Rightarrow \angle A=120^o$ , vì

$\angle A>\angle B>\angle C$

$0=\sin 5A+\sin 5B+\sin 5C=2\sin \frac{5(A+B)}{2}\cos \frac{5(A-B)}{2}+2\sin \frac{5C}{2}\cos \frac{5C}{2}$

$=2\cos \frac{5C}{2}\left(\cos \frac{5(A-B)}{2}+\cos \frac{5C}{2} \right)$

$\Rightarrow \cos \frac{5A}{2}\cos \frac{5B}{2}\cos \frac{5C}{2}=0$

$\Rightarrow \left[\begin{array}{l}\angle A=36^o\\\angle B=36^o\\\angle C=36^o\end{array}\right. \Rightarrow \angle B=36^o$ , vì

$\angle A>\angle B>\angle C$

Từ đó suy ra

$\angle A=120^o,\angle B=36^o,\angle C=24^o$

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