Speed of the cyclist, $v=27\, km / h =7.5 \,m / s$
Radius of the circular turn, $r=80 \,m$
Centripetal acceleration is given as:
$a_{e}=\frac{v^{2}}{r}$
$=\frac{(7.5)^{2}}{80}=0.7\, m / s ^{2}$
The situation is shown in the given figure
Suppose the cyclist begins cycling from point $P$ and moves toward point $Q$. At point $Q$ he applies the breaks and decelerates the speed of the bicycle by $0.5\, m / s ^{2}$
This acceleration is along the tangent at $Q$ and opposite to the direction of motion of the cyclist.
since the angle between $a_{\varepsilon}$ and $a_{ r }$ is $90^{\circ},$ the resultant acceleration $a$ is given by:
$a=\sqrt{a_{c}^{2}+a_{1}^{2}}$
$=\sqrt{(0.7)^{2}+(0.5)^{2}}$
$=\sqrt{0.74}=0.86 \,m / s ^{2}$
$\tan \theta=\frac{a_{c}}{a_{T}}$
Where $\theta$ is the angle of the resultant with the direction of velocity
$\tan \theta=\frac{0.7}{0.5}=1.4$
$\theta=\tan ^{-1}(1.4)$
$=54.46^{\circ}$