Cho tam giác $ABC$ có hệ thức :  $S=\frac{ 1}{ 4}a^2. \sin 2B $  thì $\Delta ABC$ vuông.
t thấy khó mà –  cuungonghinh 02-11-12 09:38 PM
bài này cũng dễ :D –  banhquykeomut 02-11-12 09:14 PM
Ta có:
      $S=\frac{1}{ 4}a^2.\sin 2B$
 $\Leftrightarrow \frac{1}{2}ac\sin B=\frac{1}{2}a^2\sin B\cos B$
 $\Leftrightarrow c=a.\frac{a^2+c^2-b^2}{2ac}$
$\Leftrightarrow a^2=b^2+c^2$
 $\Leftrightarrow \Delta ABC$ vuông tại $A$.
bài giải đc đó –  nadlks297 31-10-12 10:44 PM
em xem qua cái đã, cảm ơn mấy anh nhé –  ranganh 25-10-12 10:59 PM

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