Răspunsuri

2014-08-25T11:38:23+03:00
ΔABC, mas<A=90
sin<B= 2/3                  => cos <B= 1/2/3= 1* 3/2= 3/2

cos<B=3/2
2 4 2
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2014-08-25T13:08:34+03:00
Avem ΔABC dreptunghic in A  (<A = 90°)

sin <B = b/a  = 2/3
cos <B = c/a
Aplicam teorema lui Pitagora:
b² + c² = a²

 \frac{b^{2}+c^{2}}{a^{2}}=1<=> \frac{b^{2}}{a^{2}}+\frac{c^{2}}{a^{2}}=1 \\ \frac{c^{2}}{a^{2}}=1-\frac{b^{2}}{a^{2}}=> \frac{c}{a}= \sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{2^{2}}{3^{2}}} = \\ = \sqrt{1-\frac{4}{9}}= \sqrt{ \frac{5}{9}}= \frac{ \sqrt{5} }{3}

=>  cos <B = √5 / 3


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