Question

preciso de ajuda nesse exercício!

Answer 1

a)

$\\cateto\ oposto: \ 2 \ \ \ \ \ \ hipotenusa : \ 7 \ \ \ \ \ \  \hat{a}ngulo \ C \\\\Sen\ C=\frac{cateto\ oposto}{hipotenusa}\\\\Sen\ C=\frac{2}{7}$

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b)

$\\a^2=b^2+c^2\\\\a^2=11^2+60^2\\\\a^2=121+3600\\\\a^2=3721\\\\a=\sqrt{3721}\\\\a=61\\------------\\cateto\ oposto: \ 11 \ \ \ \ \ \ hipotenusa : \ 61 \ \ \ \ \ \  \hat{a}ngulo: \ B \\\\Sen\ B=\frac{cateto\ oposto}{hipotenusa}\\\\Sen\ B=\frac{11}{61}$

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c)

$a^2=b^2+c^2\\a^2=5^2+4^2\\a^2=25+16\\a^2=41\\a=\sqrt{41}\\------------------\\\\cateto\ oposto: \ 5 \ \ \ \ \ \ hipotenusa : \ \sqrt{41} \ \ \ \ \ \  \hat{a}ngulo: \ A \\\\Sen\ A=\frac{cateto\ oposto}{hipotenusa}\\\\Sen\ A=\frac{5}{\sqrt{41}}\\\\Sen\ A=\frac{5}{\sqrt{41}}\cdot\frac{\sqrt{41}}{\sqrt{41}}=\frac{5\sqrt{41}}{\sqrt{(41)^2}}=\frac{5\sqrt{41}}{41}$