Toán 9

P

pinkylun

a) Có:

$AH^2=BH.HC$

$=>AH=\sqrt{16.9}=12$

$=>S_{ABC}=\dfrac{1}{2}AH.BC=\dfrac{1}{2}.12.(16+9)=150cm^2$

$\ \tan B=\dfrac{AH}{BH}=\dfrac{3}{4}$ :D

$=>cotB=\dfrac{4}{3}$

$AB=\sqrt{16^2+12^2}=20$

$=>sinB=...;cosB=...$

$sin C=\dfrac{20}{16+9}=\dfrac{4}{5}$

$=>\hat{C}=\ \sin^{-1} (\dfrac{4}{5})$~$53^o8'$
 
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