Question

A figura a seguir representa um △ ABC, equilátero, com lado medindo 1 cm e uma altura BH.

Answer 1

$AC=AB=BC=1 cm\\BAH=BCH\\BH=h\\\\BC^{2}=BH^{2}+HC^{2} => BC^{2}=h^{2}+HC^{2}\\h^2=BC^2-HC^2 +> h=\sqrt{BC^2-HC^2}\\h=\sqrt{1^2-\frac{1}{2} ^2}=\sqrt{1-\frac{1}{4} }=\sqrt{\frac{3}{4} } =\frac{\sqrt{3} }{2}\\\\sen 60=\frac{h}{BC}=\frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3} }{2} \\cos60=\frac{0,5}{1} =1/2\\tg60=sen/cos=\sqrt{3}\\\\(b) \sqrt{3}/2, \sqrt{3}/2, 1/2, \sqrt{3}$