Here are 2 ways you could do this.

**Method 1: ...almost feels like cheating ;)**

If #tan(2b)=119/120#

use your calculator (or a spreadsheet) to evaluate:

#color(white)("XXX")2b=arctan(119/120)~~0.7812140874#

which implies (after dividing by 2)

#color(white)("XXX")b=0.3906070437#

Then

use your calculator again to find

#color(white)("XXX")cos(b)=cos(0.3906070437)~~0.9246780985#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

**Method 2: possibly more virtuous (???)**

If #tan(2b)=119/120#

then we can think of angle #(2b)# as being the angle of a right triangle in standard position with the horizontal (#x#) component equal to #120# and the vertical (#y#) component equal to #119#.

The hypotenuse would be equal to #sqrt(120^2+119^2)=169#

(and, yes, I did use a calculator to discover this).

#cos(2b)=("horizontal")/("hypotenuse")=120/169#

Then, if we remember the double angle formula:

#color(white)("XXX")cos(2b)=2cos^2(b)-1#

we can re-arrange the terms to get

#color(white)("XXX")cos(b)=sqrt((cos(2b)+1)/2)#

#color(white)("XXXXXX")=sqrt((120/169+1)/2)#

#color(white)("XXXXXX")=sqrt((289/169)/2)#

#color(white)("XXXXXX")=sqrt(289/338)#

and with the aid of my calculator (again)

#color(white)("XXXXXX")~~0.924680985#