Diketahui segitiga ABC siku-siku di B dan sin A = 1/3 nilai tan A = .... * 4 poin A.1/5 √2 B.2/5 √2 C.1/3 √2 D.3/5 √2 E.2/3 √2
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Penjelasan dengan langkah-langkah:
[tex]\boxed{sin(A) = \frac{depan}{miring } = \frac{ 1 }{3} }[/tex]
[tex] \boxed{cos(A) = \frac{samping}{miring} = \frac{ \sqrt{ {(3)}^{2} - {(1)}^{2} } }{3} = \frac{ \sqrt{8} }{3} = \frac{2 \sqrt{2} }{3} }[/tex]
[tex]tan(A) = \frac{sin(A) }{cos(A) } = \frac{ \frac{1}{3} }{ \frac{2 \sqrt{2} }{3} } = \frac{1}{2 \sqrt{2} } [/tex]
[tex]tan(A) = \frac{1}{2 \sqrt{2} } \times \frac{2 \sqrt{2} }{2 \sqrt{2} } = \frac{2 \sqrt{2} }{8} [/tex]
[tex]\boxed{tan(A) = \frac{2 \sqrt{2} }{8} = \frac{1}{4} \sqrt{2} }[/tex]
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