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

Răspuns :

Din teorema cosinusului [tex]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}= [/tex]
[tex] \sqrt{1- \frac{9}{25} }= \sqrt{ \frac{16}{25} }= \frac{4}{5} [/tex]