Question

Lượng giác 

Cho tam giác ABC,AB=10cm,AC=15cm,Â=80⁰ A)Tính BC B)Tính góc B,C

Answer 1

a: Xét ΔABC có

\(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

=>\(10^2+15^2-BC^2=2\cdot10\cdot15\cdot cos80\)

=>\(BC^2=325-300\cdot cos80\)

=>\(BC\simeq16,52\left(cm\right)\)

b: Xét ΔABC có \(\dfrac{BC}{sinA}=\dfrac{AC}{sinB}=\dfrac{AB}{sinC}\)

=>\(\dfrac{15}{sinB}=\dfrac{10}{sinC}=\dfrac{16.52}{sin80}\)

=>\(\left\{{}\begin{matrix}sinB=\dfrac{15\cdot sin80}{16.52}\simeq0.89\\sinC\simeq0.6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\widehat{B}\simeq63^0\\\widehat{C}\simeq57^0\end{matrix}\right.\)