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pinkylun

a)Xét $\triangle{ABC}$ có:

$BC^2=26^2=676$

$AC^2+AB^2=10^2+24^2=676$

$=>BC^2=AB^2+AC^2 $=>đpcm

b)$ \sin B=\dfrac{AC}{BC}=\dfrac{24}{26}=\dfrac{12}{13}$

$\sin C=\dfrac{AB}{BC}=\dfrac{10}{26}=\dfrac{5}{13}$

c) $ \sin B=\dfrac{AH}{AB}$

$=>AH=\sin B .10=\dfrac{120}{13}$~$9cm$

$=>BH=\sqrt{10^2-(\dfrac{120}{13})^2}=\dfrac{50}{13}$~$4cm$

$=>CH$~$26-4$~$22cm$
 
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