Ok in question 52 I know how to get a)the centripetal accelration of a rider and

Question

Ok in question 52 I know how to get
a)the centripetal accelration of a rider and what force the seatexerts on a 40.0kg rider. b)the lowest point of the rider c)the highest point of the rider and d)the force the seat exerts on the rider when the rider ishalfway between the top and bottom What I do not understand is why tan=r/g turns into=tan^-1r/g. How am I supposed to know thattan turns into =tan^-1? Ok in question 52 I know how to get
a)the centripetal accelration of a rider and what force the seatexerts on a 40.0kg rider. b)the lowest point of the rider c)the highest point of the rider and d)the force the seat exerts on the rider when the rider ishalfway between the top and bottom What I do not understand is why tan=r/g turns into=tan^-1r/g. How am I supposed to know thattan turns into =tan^-1?

Explanation / Answer

Here tan has not changed its value butwe have to solvefor angle . Here tan = ar/g   then solving for by mathematics and triginometry we get =tan-1(ar/g) then solving for by mathematics and triginometry we get =tan-1(ar/g)

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