Question

Answer 1

Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{24^2+BC^2-30^2}{2\cdot24\cdot BC}=cos60=\dfrac{1}{2}\)

=>\(BC^2-324=24\cdot BC\)

=>\(BC^2-24\cdot BC-324=0\)

=>\(BC^2-24BC+144-468=0\)

=>\(\left(BC-12\right)^2=468\)

=>\(\left[{}\begin{matrix}BC-12=\sqrt{468}=6\sqrt{13}\\BC-12=-6\sqrt{13}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}BC=6\sqrt{13}+12\left(nhận\right)\\BC=12-6\sqrt{13}=6\sqrt{4}-6\sqrt{13}< 0\left(loại\right)\end{matrix}\right.\)

Vậy: \(BC=6\sqrt{3}+12\simeq33,63\left(cm\right)\)

Xét ΔABC có \(\dfrac{AB}{sinC}=\dfrac{AC}{sinB}\)

=>\(\dfrac{30}{sin60}=\dfrac{24}{sinC}\)

=>\(sinC=24\cdot\dfrac{sin60}{30}=\dfrac{2\sqrt{3}}{5}\)

=>\(\widehat{C}\simeq44^0\)