Question

Giải trắc nghiệm nhé,chi tiết giúp tớ với ạ !

Answer 1

Câu 1: Xét ΔABC có \(cosA=\dfrac{b^2+c^2-a^2}{2\cdot b\cdot c}\)

=>\(b^2+c^2-a^2=2\cdot b\cdot c\cdot cosA\)

=>\(a^2=b^2+c^2-2bc\cdot cosA\)

=>Chọn B

Câu 2: Xét ΔABC có 

\(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)

=>\(\dfrac{8^2+10^2-AB^2}{2\cdot8\cdot10}=cos60=\dfrac{1}{2}\)

=>\(164-AB^2=80\)

=>\(AB^2=164-80=84\)

=>\(AB=\sqrt{84}=2\sqrt{21}\)

=>Chọn D

Câu 3:

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

=>\(\dfrac{6^2+8^2-BC^2}{2\cdot6\cdot8}=cos60=\dfrac{1}{2}\)

=>\(100-BC^2=48\)

=>\(BC^2=100-48=52\)

=>\(BC=\sqrt{52}=2\sqrt{13}\)

=>Chọn A

Câu 4: Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{8^2+5^2-AC^2}{2\cdot8\cdot5}=cos60=\dfrac{1}{2}\)

=>\(64+25-AC^2=40\)

=>\(AC^2=49=7^2\)

=>AC=7

=>Chọn  A

Câu 5: Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{9^2+8^2-AC^2}{2\cdot9\cdot8}=cos60=\dfrac{1}{2}\)

=>\(81+64-AC^2=72\)

=>\(AC^2=81+64-72=64+9=73\)

=>\(AC=\sqrt{73}\)

=>Chọn A

Câu 6: Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

=>\(\dfrac{2^2+1^2-BC^2}{2\cdot2\cdot1}=cos60=\dfrac{1}{2}\)

=>\(5-BC^2=\dfrac{4}{2}=2\)

=>\(BC^2=3\)

=>\(BC=\sqrt{3}\)

=>Chọn C

Câu 7; Xét ΔABC có \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)

=>\(\dfrac{8^2+3^2-AC^2}{2\cdot8\cdot3}=cos60=\dfrac{1}{2}\)

=>\(64+9-AC^2=8\cdot3=24\)

=>\(AC^2=73-24=49\)

=>AC=7

=>Chọn C

Câu 8: Xét ΔABC có \(cosC=\dfrac{CA^2+CB^2-AB^2}{2\cdot CA\cdot CB}\)

=>\(\dfrac{\left(\sqrt{3}\right)^2+2^2-AB^2}{2\cdot\sqrt{3}\cdot2}=cos150=\dfrac{-\sqrt{3}}{2}\)

=>\(7-AB^2=4\sqrt{3}\cdot\dfrac{-\sqrt{3}}{2}=2\cdot\left(-3\right)=-6\)

=>\(AB^2=13\)

=>\(AB=\sqrt{13}\)

=>Chọn A

Câu 9: Xét ΔABC có \(cosA=\dfrac{BA^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

=>\(\dfrac{7^2+5^2-BC^2}{2\cdot7\cdot5}=\dfrac{4}{5}\)

=>\(49+25-BC^2=2\cdot7\cdot5\cdot\dfrac{4}{5}=2\cdot7\cdot4=56\)

=>\(BC^2=49+25-56=18\)

=>\(BC=3\sqrt{2}\)

=>Chọn A