Answer:
cosΘ = [tex]\frac{\sqrt{51} }{10}[/tex]
Step-by-step explanation:
Using the trigonometric identity
sin²x + cos²x = 1 ⇒ cosx = ± [tex]\sqrt{1-sin^2x}[/tex]
Given
sinΘ = [tex]\frac{7}{10}[/tex], then
cosΘ = ± [tex]\sqrt{1-(7/10)^2}[/tex] = ± [tex]\sqrt{1-\frac{49}{100} }[/tex] = ± [tex]\sqrt{\frac{51}{100} }[/tex] = ± [tex]\frac{\sqrt{51} }{10}[/tex]
Since Θ is in first quadrant then cosΘ > 0
cosΘ = [tex]\frac{\sqrt{51} }{10}[/tex]
