Question

If `T=50+50cos((pit)/6)`, express `t` as a function of `T`?

Answer 1

`t=6/piarccos(T/50-1)`

Explanation:

As `T=50+50cos((pit)/6)`

`cos((pit)/6)=(T-50)/50`

and `(pit)/6=arccos(T/50-1)`

i.e. `t=6/piarccos(T/50-1)`