Question

Se considera triunghiul ABC,cu AB=10,AC=10,si BC=12.Aratati ca sinB=4/5

Answer 1

Din teorema cosinusului $cosB= \frac{AB^2+BC^2-AC^2}{2AC*BC}= \frac{100+144-100}{2*10*12}= \frac{3}{5};iar,sinB= \sqrt{1-cos^2B}=$
$\sqrt{1- \frac{9}{25} }= \sqrt{ \frac{16}{25} }= \frac{4}{5}$