Áp dụng định lí Pi-ta-go vào tam giác vuông ABC, ta có:
\(B{C^2} = A{B^2} + A{C^2}\)
\(\eqalign{
& \Rightarrow A{B^2} = B{C^2} - A{C^2} \cr
& = B{C^2} - \dfrac{{B{C^2}}}{4} = \dfrac{{3B{C^2}}}{4} \cr
& \Rightarrow AB = \dfrac{{BC\sqrt 3 }}{ 2} \cr} \)
Vậy: \(\sin \widehat B = \dfrac{{AC}}{{BC}} = \dfrac{{\dfrac{1}{2}BC}}{{BC}} = \dfrac{1}{2}\)
\({\rm{cos}}\widehat B = \dfrac{{AB}}{{BC}} = \dfrac{{\dfrac{{\sqrt 3 }}{2}BC}}{{BC}} = \dfrac{{\sqrt 3 }}{2}\)
\(tg\widehat B = \dfrac{{AC}}{{AB}} = \dfrac{{\dfrac{1}{2}BC}}{{\dfrac{{\sqrt 3 }}{2}BC}} = \dfrac{{\sqrt 3 }}{ 3}\)
\(\cot g\widehat B = \dfrac{1}{{tgB}} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{3}}} = \sqrt 3 \)
