Answer:
cosΘ = [tex]\frac{11}{12}[/tex]
Step-by-step explanation:
Using the trigonometric identity
sin²x + cos²x = 1 ⇒ cosx = ± [tex]\sqrt{1-sin^2x}[/tex]
Given
sinΘ = [tex]\frac{\sqrt{23} }{12}[/tex], then
cosΘ = ± [tex]\sqrt{1-(\frac{\sqrt{23} }{12} }[/tex])² = ± [tex]\sqrt{1-\frac{23}{144} }[/tex] = ± [tex]\sqrt{\frac{121}{144} }[/tex] = ± [tex]\frac{11}{12}[/tex]
Since Θ is in first quadrant then cosΘ > 0
cosΘ = [tex]\frac{11}{12}[/tex]
