1. Problem: Check if there exists an acute angle $\alpha$ such that $\sin\alpha = \frac{3\sqrt{34}}{34}$ and $\tan\alpha = \frac{4}{5}$. 2. Recall the identity relating sine and tangent: $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}$$ 3. From $\tan\alpha = \frac{4}{5}$, express $\cos\alpha$: $$\cos\alpha = \frac{\sin\alpha}{\tan\alpha} = \frac{\frac{3\sqrt{34}}{34}}{\frac{4}{5}} = \frac{3\sqrt{34}}{34} \times \frac{5}{4} = \frac{15\sqrt{34}}{136}$$ 4. Check if $\sin^2\alpha + \cos^2\alpha = 1$: $$\left(\frac{3\sqrt{34}}{34}\right)^2 + \left(\frac{15\sqrt{34}}{136}\right)^2 = \frac{9 \times 34}{34^2} + \frac{225 \times 34}{136^2} = \frac{306}{1156} + \frac{7650}{18496}$$ 5. Simplify denominators: $$1156 = 34^2, \quad 18496 = 136^2$$ 6. Convert to common denominator $18496$: $$\frac{306}{1156} = \frac{306 \times 16}{1156 \times 16} = \frac{4896}{18496}$$ 7. Sum: $$\frac{4896}{18496} + \frac{7650}{18496} = \frac{12546}{18496} \neq 1$$ 8. Since $\sin^2\alpha + \cos^2\alpha \neq 1$, no such acute angle $\alpha$ exists. Final answer: No, there is no acute angle $\alpha$ satisfying both conditions.