3:
a: \(cosB=\dfrac{BA^2+BC^2-AC^2}{2\cdot BA\cdot BC}\)
=>\(5^2+6^2-AC=2\cdot5\cdot6\cdot\dfrac{\sqrt{2}}{2}=30\sqrt{2}\)
=>\(AC\simeq4.31\left(cm\right)\)
BA/sinC=BC/sinA=AC/sinB
=>5/sinC=6/sinA=4,31/sin45
=>góc C=55 độ; góc A=80 độ
b: S ABC=1/2*BA*BC*sinC=1/2*5*6*sin45=15/2*căn 2
AC/sinB=2R
=>R=3,05(cm)
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