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Derive an expression for the angle at which a cyclist will have to make with vertical while taking a circular turn.

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The forces acting on the cycle and cyclists are weight of the cycle and cyclist together = mg vertically downwards and the reaction by the ground on the cycle tyre at an angle θ to the vertical. Resolve this reaction R in Vertical component - Rcosθ and Horizontal component Rsinθ. Equating...
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The forces acting on the cycle and cyclists are weight of the cycle and cyclist together = mg vertically downwards

and the reaction by the ground on the cycle tyre at an angle θ to the vertical. Resolve this reaction R in Vertical component - Rcosθ and Horizontal component Rsinθ. Equating the vertical forces Rcosθ = mg and the horizontal force is nothing but the centripetal force Rsinθ = mv2/r; 

Divide second equation by first Rsinθ/Rcosθ = v2/rg; tanθ = v2/rg or θ = tan-1(v2/rg)

Please remember we have ignored the frictional force between the tyre and the ground here

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