Step-by-Step Solution
Key Concept: Use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ to find $\cos A$, and then use the definition $\tan A = \dfrac{\sin A}{\cos A}$.
1. Given $\sin A = \dfrac{3}{4}$.\
2. Apply the identity $\sin^2 A + \cos^2 A = 1$:\
$$\left(\dfrac{3}{4}\right)^2 + \cos^2 A = 1$$\
$$\dfrac{9}{16} + \cos^2 A = 1$$\
$$\cos^2 A = 1 - \dfrac{9}{16} = \dfrac{7}{16}.$$\
3. Since $A$ is an acute angle (implied by $\sin A = \frac34 >0$), $\cos A$ is positive. Hence\
$$\cos A = \sqrt{\dfrac{7}{16}} = \dfrac{\sqrt{7}}{4}.$$\
4. Now compute $\tan A$ using $\tan A = \dfrac{\sin A}{\cos A}$:\
$$\tan A = \dfrac{\dfrac{3}{4}}{\dfrac{\sqrt{7}}{4}} = \dfrac{3}{\sqrt{7}} = \dfrac{3\sqrt{7}}{7}.$$
Correct Answer: cos A = \dfrac{\sqrt{7}}{4}, \; tan A = \dfrac{3\sqrt{7}}{7}