Question

Answer 1

a) Ta có:

\(\widehat{C}=90^0-\widehat{A}=90^0-58^0=32^0\)

\(sinB=\dfrac{AC}{BC}\)

\(\Rightarrow AC=BC.sinB=72.sin58^0\approx61,1\left(cm\right)\)

\(sinC=\dfrac{AB}{BC}\)

\(\Rightarrow AB=BC.sinC=72.sin32^0\approx38,2\left(cm\right)\)

b) Ta có:

\(\widehat{C}=90^0-\widehat{B}=90^0-48^0=42^0\)

\(sinB=\dfrac{AC}{BC}\)

\(\Rightarrow BC=\dfrac{AC}{sinB}=\dfrac{20}{sin48^0}=\approx26,9\left(cm\right)\)

\(sinC=\dfrac{AB}{BC}\)

\(\Rightarrow AB=BC.sinC=26,9.sin42^0\approx18\left(cm\right)\)

c) Ta có:

\(\widehat{B}=90^0-\widehat{C}=90^0-30^0=60^0\)

\(sinB=\dfrac{AC}{BC}\)

\(\Rightarrow BC=\dfrac{AC}{sinB}=\dfrac{15}{sin60^0}=10\sqrt{3}\left(cm\right)\)

\(sinC=\dfrac{AB}{BC}\)

\(\Rightarrow AB=BC.sinC=10\sqrt{3}.sin30^0=5\sqrt{3}\left(cm\right)\)

d) Ta có:

\(\widehat{B}=90^0-\widehat{C}=90^0-18^0=72^0\)

\(sinB=\dfrac{AC}{BC}\)

\(\Rightarrow BC=\dfrac{AC}{sinB}=\dfrac{21}{sin72^0}\approx22,1\left(cm\right)\)

\(sinC=\dfrac{AB}{BC}\)

\(\Rightarrow AB=BC.sinC=22,1.sin18^0\approx6,8\left(cm\right)\)