Answer: [tex]\sin \theta=\frac{-45}{53}[/tex]
Step-by-step explanation:
Since we have given that
[tex]\cos\theta=\frac{28}{53}[/tex]
And we know that θ is in the Fourth Quadrant.
So, Except cosθ and sec θ, all trigonometric ratios will be negative.
As we know the "Trigonometric Identity":
[tex]\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}[/tex]
It must be negative due to its presence in Fourth quadrant.
Hence, [tex]\sin \theta=\frac{-45}{53}[/tex]
