Bài tập 6.33 trang 190 SBT Toán 10

\(\left\{ \begin{array}{l}
\widehat {ABD} = \widehat {ADB}\\
\widehat {ABD} = \widehat {BDC}
\end{array} \right. \Rightarrow \widehat {BDC} = \widehat {ADB}\)
Suy ra \(\widehat {BAD} = \pi  - 2\widehat {BDC}\)

Từ đó ta có:

\(\begin{array}{*{20}{l}}
{\tan \widehat {BAD} =  - \tan 2\widehat {BDC} =  - \frac{{2\tan \widehat {BCD}}}{{1 - {{\tan }^2}\widehat {BDC}}}}\\
{ =  - \frac{{2.\frac{3}{4}}}{{1 - \frac{9}{{16}}}} =  - \frac{{24}}{7}}
\end{array}\)

Vì \(\frac{\pi }{2} < \widehat {BAD} < \pi \) nên \(\cos \widehat {BAD} < 0\). Do đó

 \(\cos \widehat {BAD} =  - \frac{1}{{\sqrt {1 + {{\tan }^2}BAD} }} =  - \frac{1}{{\sqrt {1 + \frac{{576}}{{40}}} }} =  - \frac{7}{{25}}\)

\(\sin \widehat {BAD} = \cos \widehat {BAD}.\tan \widehat {BAD} = \frac{{ - 7}}{{25}}.\frac{{ - 24}}{7} = \frac{{24}}{{25}}\)

-- Mod Toán 10 HỌC247