Giải nhanh hộ em với

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Giải nhanh hộ em với
giai-nhanh-ho-em-voi

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Doris 6 years 2020-11-24T12:51:24+00:00 1 Answers 77 views 0

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  1. Delwyn
    0
    2020-11-24T12:52:59+00:00
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    Giải thích các bước giải:

     Ta có:

    \(\begin{array}{l}
    a,\\
    {\sin ^2}B + {\cos ^2}B = 1\\
     \Leftrightarrow {\sin ^2}B + 0,{5^2} = 1\\
     \Leftrightarrow {\sin ^2}B = \dfrac{3}{4}\\
    0^\circ  < \widehat B < 90^\circ  \Rightarrow \sin B > 0 \Rightarrow \sin B = \dfrac{{\sqrt 3 }}{2}\\
    \tan B = \dfrac{{\sin B}}{{\cos B}} = \sqrt 3 \\
    \cot B = \dfrac{{\cos B}}{{\sin B}} = \dfrac{1}{{\sqrt 3 }}\\
    \widehat B + \widehat C = 90^\circ  \Rightarrow \left\{ \begin{array}{l}
    \sin C = \cos B = \dfrac{1}{2}\\
    \cos B = \sin B = \dfrac{{\sqrt 3 }}{2}
    \end{array} \right.\\
    \tan C = \dfrac{{\sin C}}{{\cos C}} = \dfrac{1}{{\sqrt 3 }}\\
    \cot C = \dfrac{{\cos C}}{{\sin C}} = \sqrt 3 \\
    b,\\
    \tan C = 1,2 \Rightarrow \dfrac{{\sin C}}{{\cos C}} = 1,2 \Rightarrow \sin C = \dfrac{6}{5}\cos C\\
    {\sin ^2}C + {\cos ^2}C = 1\\
     \Leftrightarrow {\left( {\dfrac{6}{5}\cos C} \right)^2} + {\cos ^2}C = 1\\
     \Leftrightarrow \dfrac{{61}}{{25}}{\cos ^2}C = 1\\
     \Rightarrow {\cos ^2}C = \dfrac{{25}}{{61}}\\
     \Rightarrow \cos C = \dfrac{5}{{\sqrt {61} }}\\
     \Rightarrow \sin C = \dfrac{6}{5}\cos C = \dfrac{6}{{\sqrt {61} }}\\
    \cot C = \dfrac{{\cos C}}{{\sin C}} = \dfrac{5}{6}\\
    \widehat B + \widehat C = 90^\circ  \Rightarrow \left\{ \begin{array}{l}
    \sin C = \cos B = \dfrac{6}{{\sqrt {61} }}\\
    \cos B = \sin B = \dfrac{5}{{\sqrt {61} }}
    \end{array} \right.\\
    \tan B = \dfrac{{\sin B}}{{\cos B}} = \dfrac{5}{6}\\
    \cot B = \dfrac{{\cos B}}{{\sin B}} = \dfrac{6}{5}
    \end{array}\)

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