Đáp án:
B1: $A = 3;B = \dfrac{3}{2};C = 1$
B2: ${\widehat B = {{47}^0};AC \approx 5,36;BC = 7,33}$
Giải thích các bước giải:
Bài 1:
$\begin{array}{l}
a)A = 2 + 3\sin {43^0} + \tan {45^0} - 3\cos {47^0}\\
= 2 + 3\left( {\sin {{43}^0} - \cos {{47}^0}} \right) + \tan {45^0}\\
= 2 + 3\left( {\cos {{47}^0} - \cos {{47}^0}} \right) + 1\\
= 2 + 1\\
= 3\\
b)B = {\cos ^2}{37^0} + {\sin ^2}{45^0} + {\cos ^2}{53^0}\\
= \left( {{{\cos }^2}{{37}^0} + {{\cos }^2}{{53}^0}} \right) + {\sin ^2}{45^0}\\
= \left( {{{\sin }^2}{{53}^0} + {{\cos }^2}{{53}^0}} \right) + {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2}\\
= 1 + \dfrac{1}{2}\\
= \dfrac{3}{2}\\
c)C = \dfrac{{\sin {{42}^0}}}{{\cos {{48}^0}}} + \dfrac{1}{{\tan {{45}^0}}} - \cot {21^0}.\cot {69^0}\\
= \dfrac{{\cos {{48}^0}}}{{\cos {{48}^0}}} + \dfrac{1}{1} - \tan {69^0}.\cot {69^0}\\
= 1 + 1 - 1\\
= 1
\end{array}$
Vậy $A = 3;B = \dfrac{3}{2};C = 1$
Bài 2:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AB = 5;\widehat C = {43^0}\\
\Rightarrow \left\{ \begin{array}{l}
\widehat B = {90^0} - \widehat C = {47^0}\\
AC = AB.\tan B \approx 5,36\\
BC = \dfrac{{AB}}{{\cos B}} = 7,33
\end{array} \right.
\end{array}$
Vậy ${\widehat B = {{47}^0};AC \approx 5,36;BC = 7,33}$
