Answer: The answer is [tex]\cos \theta=\dfrac{\sqrt{7}}{4}.[/tex]
Step-by-step explanation: Given that
[tex]\sin \theta=\dfrac{3}{4}[/tex] and [tex]\theta[/tex] is in Quadrant I.
We are to find the value of [tex]\cos \theta.[/tex]
We have the following trigonometric identity :
[tex]\sin^2\theta+\cos^2\theta=1\\\\\Rightarrow \cos^2\theta=1-\sin^2\theta\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\sin^2\theta}\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\left(\dfrac{3}{4}\right)^2}\\\\\\\Rightarrow \cos \theta=\pm\sqrt{1-\dfrac{9}{16}}\\\\\\\Rightarrow \cos \theta=\pm\sqrt{\dfrac{7}{16}}\\\\\\\Rightarrow \cos \theta=\pm\dfrac{\sqrt{7}}{4}.[/tex]
Since [tex]\theta[/tex] lies in the first quadrant, so cosine of [tex]\theta[/tex] will be positive.
Thus,
[tex]\cos \theta=\dfrac{\sqrt{7}}{4}.[/tex]
