Lượng giác 10

L

lp_qt

$b.cosC+c.cosB=b.\dfrac{a^2+b^2-c^2}{2ab}+c.\dfrac{a^2+c^2-b^2}{2ac}$

$=\dfrac{a^2+b^2-c^2+a^2+c^2-b^2}{2a}=a$

$b.sinC.(b.cosC+c.cosB)=b.sinC.a=2.S_{ABC}$

\Rightarrow $S_{ABC}=10$
 
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