A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 metres/sec. The combined mass of the boy and the cycle is 90kg . The angle that the cycle makes with the vertical so that it may not fall is (g= 9.8 m / sec²)

To find the angle that the cycle makes with the vertical so that the cyclist does not fall while turning, we can analyze the forces acting on the cycle and the rider system in a circular path.

When a cyclist moves in a circular path of radius $r$ with a constant speed $v$, they must lean inwards towards the center of the curve at an angle $\theta$ with the vertical to provide the necessary centripetal force and maintain balance.
The relation between the angle $\theta$ with the vertical, the speed $v$, the radius of the circle $r$, and the acceleration due to gravity $g$ is given by the formula:

$\tan \left(\theta \right)=\frac{v^2}{rg}$

Let us write down the given parameters from the problem:
Radius of the circular path, $r=20\text{m}$
Speed of the cycle, $v=20\text{m/s}$
Combined mass of the boy and the cycle, $m=90\text{kg}$ (Note: the mass cancels out in the balance equation and is not required for finding the angle)
Acceleration due to gravity, $g=9.8{\text{m/s}}^2$

Now, let's substitute the given values into the formula:

$\tan \left(\theta \right)=\frac{{\left(20\right)}^2}{20\times 9.8}$

Simplify the numerator and denominator:

$\tan \left(\theta \right)=\frac{400}{196}$

Calculating the value of the fraction:

$\tan \left(\theta \right)\approx 2.0408$

To find the angle $\theta$, we take the inverse tangent (arctan) of both sides:

$\theta ={\tan }^{-1}\left(2.0408\right)$

$\theta \approx 63.90°$

Therefore, the angle that the cycle makes with the vertical so that it may not fall is approximately 63.90°.