Net acceleration is due to braking and centripetal accelerationDue to Braking,\\\\(a_T= 0.5 m/s^2\\\\)Speed of the cyclist, \\\\(v = 27 km/h = 7.5 m/s\\\\)Radius of the circular turn, \\\\(r=80m\\\\)Centripetal acceleration is given as:\\\\(a_c= \\dfrac{V^2}{r}\\\\) \\\\(=(7.5 ^2)/80= 0.70 m/s^2\\\\)Since the angle between \\\\(a_c\\\\) and \\\\(a_T\\\\) is 900, the resultant acceleration a is given by:\\\\(a= (a_c^2+a_T^2)^{1/2} \\\\)\\\\(a=(0.7^2+ 0.5^2)^{1/2}= 0.86 m/s^2\\\\)\\\\(tan \ heta= \\dfrac{a_c}{a_T}\\\\)where \\\\(\ heta \\\\) is the angle of the resultant with the direction of the velocity. \\\\(tan \ heta=\\dfrac{0.7}{0.5}= 1.4 \\\\)\\\\(\ heta = tan^{-1}(1.4) = 54.56^0\\\\)