Question

Answer 1

a: Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(\widehat{C}=180^0-60^0-40^0=80^0\)

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

=>\(\dfrac{BC}{sin60}=\dfrac{AC}{sin40}=\dfrac{14}{sin80}\)

=>\(BC=14\cdot\dfrac{sin60}{sin80}\simeq12,31;AC=\dfrac{14}{sin80}\cdot sin40\simeq9,14\)

b: Xét ΔABC có \(\widehat{A}+\widehat{C}+\widehat{B}=180^0\)

=>\(\widehat{B}=180^0-30^0-75^0=75^0\)

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

=>\(\dfrac{4.5}{sin75}=\dfrac{AB}{sin75}=\dfrac{BC}{sin30}\)

=>\(AB=4,5;BC=4.5\cdot\dfrac{sin30}{sin75}=\dfrac{9\sqrt{6}-9\sqrt{2}}{4}\)