\\\\(\ extbf{Step 1: Drawing FBD}\\\\) \\\\(\ extbf{[Refer Fig. 1]}\\\\)\\\\(\ extbf{Step 2: Finding Resultant Force}\\\\) \\\\(\ extbf{[Refer Fig. 2]}\\\\)\\\\(\\large \ ext{Magnitude of Resultant:}\\\\)Since \\\\(F_1\\\\) and \\\\(F_2\\\\) are perpendicular\\\\(\ herefore \\\\) Magnitude of Resultant force on body is given by \\\\(R = \\sqrt{F_1^2 + F_2^2}\\\\) \\\\(=\\sqrt{8^2+6^2} N\\\\) \\\\(=10N\\\\)\\\\(\\large \ ext{Direction of Resultant:}\\\\) \\\\(\ heta\\ = \ an^{-1} \\dfrac{F_2}{F_1}\\\\) \\\\(\ extbf{[Refer Fig. 2]}\\\\) \\\\(=\ an^{-1} \\dfrac{6}{8}\\\\) \\\\(=36.9^o\\\\) \\\\(\\approx 37^o\\\\) \\\\(\ extbf{Step 3: Applying Newton's second Law for finding acceleration}\\\\) \\\\(\ extbf{[Refer Fig. 3]}\\\\) \\\\(F_{net} = ma\\\\) \\\\(\\Rightarrow\\\\) \\\\(10 = 5a\\\\) \\\\(\\Rightarrow\\\\) \\\\(|\\vec{a}| = 2m/s^2\\\\)Hence magnitudes of acceleration is \\\\(2m/s^2\\\\) directed at \\\\(37^o\\\\) from the 8 N force.