Question

A merry-go-round has a radius of 18 feet. If a passenger gets on a

Answer 1

Answer and Step-by-step explanation:

$Greetings!$

$Let's~answer~ your ~question!$

$\underline{\bold{It~is~given~that:}}$

$The ~radius~ of ~the ~round ~path: r = 18~ feet$

$The~ length ~of ~the ~distance ~cover ~by ~wheel : l = 38 ~feet$

$Let ~The ~angle~ of ~rotation :\phi$

$\underline{\bold{Now, ~According ~to~ question:}}$

$\therefore length~ of ~the ~arc~ of ~a ~circle ~subtending ~an ~angle~ at~ center = \phi$

$length ~of ~the ~distance ~cover ~by ~wheel =\boxed{~~ \pi \times radius \times~\frac{\phi}{180\degree^{o} }~ ~}$

$Since ~180^{o} \ = \pi ~radian$

$So,~ length~ of~ the ~distance~ cover~ by ~wheel~ \boxed{~~180^{o} \times raius \times \frac{\phi}{180^{o}} ~~}$

$\underline{\bold{i.e:}}$

$l=r \times \phi$

$Or, \phi=\frac{38~feet}{18~feet}$

$Or,~\phi =2.11^{o}$

$\bold{Thus,~the~angle~of~rotation~is=2.11^{o}}$

Answer 2

Answer:

121 degrees

Step-by-step explanation:

The "arc length" formula is s = rФ, where Ф represents the central angle in radians (not degrees).

Here r = 18 ft and s = 38 ft, and so:

                                     38 ft

s = rФ becomes Ф = ------------ = 2.111 radians

                                     18 ft

which, in degrees, is:

2.111 rad      180 deg

------------- * --------------- = 121 degrees, to the nearest degree

      1            3.142