I got this question wrong but i am unsure of the steps neede

I got this question wrong but i am unsure of the steps needed to come up with 1.89x10^8 m/s

Problem 23.08 Part A The critical angle in air for a particular type of glass is 39.0. What is the speed of light in this class glass? (c 3.00 x108 m/s) 1.97 108 m/s 1.97 x 108 m/s 1.89 x 108 m/s 1.94 x 108m/s 2.00 x 108 m/s 1.91 x 108 m/s Submit Mv Answers Give Up Incorrect, correct answer displayed

Solution

Critical angle in air for a particular type of glass C = 39 o

Relation between critical angle C and index of refraction n is n = 1/sin C

                                n = 1/sin 39

                                  = 1.589

We know index of refraction n = speed of light in vacuum / speed of light in the meterial

                                             = c / speed of light in the meterial

speed of light in the meterial = c / n

                                          =(3 x10 ⁸ m/s) / 1.589

                                          = 1.8879 x10 ⁸ m/s

                                          = 1.89 x10 ⁸ m/s        ( three significant figures)