A cyclist riding a bicycle at a speed of 14sqrt3 m/s takes a turn around a circular road of radius 20sqrt3 m without skidding. Given g = 9.8 m/s2, what is his inclination to the vertical ______.

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A cyclist riding a bicycle at a speed of 14sqrt3 m/s takes a turn around a circular road of radius 20sqrt3 m without skidding. Given g = 9.8 m/s2, what is his inclination to the vertical $60^\circ$. Explanation: v^2 = (14 sqrt 3)^2 = 142 $\times$ 3 = 196 $\times$ 3 = 588 rg = (20 sqrt 3) xx 9.8 = 20 xx 9.8 xx sqrt 3 = 196 sqrt 3 v^2/(rg) = 588/(196 sqrt 3) = $\frac{588}{196}$ xx 1/sqrt 3 = 3 xx 1/sqrt 3 = sqrt 3 theta = tan^-1 (sqrt 3) tan 60^circ = sqrt 3 $\theta$ = 60∘

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