Đáp án:
$\begin{array}{l}
a)\sin C = \dfrac{3}{5};\cos C = \dfrac{4}{5};\tan C = \dfrac{3}{4};\cot C = \dfrac{4}{3}\\
b)\tan C + \sin B = \dfrac{{31}}{{20}}
\end{array}$
Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AB = 6cm;AC = 8cm\\
\Rightarrow BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{6^2} + {8^2}} = 10cm\\
\Rightarrow \left\{ \begin{array}{l}
\sin C = \dfrac{{AB}}{{BC}} \Rightarrow \sin C = \dfrac{3}{5}\\
\cos C = \dfrac{{AC}}{{BC}} \Rightarrow \cos C = \dfrac{4}{5}\\
\tan C = \dfrac{{AB}}{{AC}} \Rightarrow \tan C = \dfrac{3}{4}\\
\cot C = \dfrac{{AC}}{{AB}} \Rightarrow \cot C = \dfrac{4}{3}
\end{array} \right.
\end{array}$
Vậy $\sin C = \dfrac{3}{5};\cos C = \dfrac{4}{5};\tan C = \dfrac{3}{4};\cot C = \dfrac{4}{3}$
b) Ta có:
$\Delta ABC;\widehat A = {90^0};AB = 6cm;AC = 8cm;BC = 10cm$
$ \Rightarrow \sin B = \dfrac{{AC}}{{BC}} = \dfrac{4}{5}$
Khi đó:
$ \tan C + \sin B = \dfrac{3}{4} + \dfrac{4}{5} = \dfrac{{31}}{{20}}$
Vậy $\tan C + \sin B = \dfrac{{31}}{{20}}$
