Answer:
cosΘ = [tex]\frac{\sqrt{3} }{2}[/tex]
Step-by-step explanation:
sin²Θ + cos²Θ = 1 ( subtract sin²Θ from both sides )
cos²Θ = 1 - sin²Θ ( take square root of both sides )
cosΘ = ± [tex]\sqrt{1-sin^20}[/tex]
= ± [tex]\sqrt{1-(\frac{1}{2})^2 }[/tex]
= ± [tex]\sqrt{1-\frac{1}{4} }[/tex]
= ± [tex]\sqrt{\frac{3}{4} }[/tex]
= ± [tex]\frac{\sqrt{3} }{2}[/tex]
since 0° < Θ < 90° , then
cosΘ = [tex]\frac{\sqrt{3} }{2}[/tex]
