How to calculate gps location based on direction and distance?
The answer in a plane you linked works rather well in a sphere provided that the distance you want to measure is tiny compared with both the radius of Earth and the distance to the nearest pole - and 2.67 chains is that tiny.

For the remaining of the answer I'll be assuming that the Earth is an sphere, which quite accurate unless you need great accuracy - that is, enough accurate for nearly anything except for surveying.

The first step is to convert your distance and direction to easting and northing distances - as the linked answer tells. 

The second step is to convert those two distances to degrees latitude and longitude, and you only need to know that one degree latitude measures $R_{Earth}\cdot\frac{\pi}{180}$ and a degree longitude measures $R_{Earth}\cdot\frac{\pi}{180}\cdot cos(latitude)$.

As a final step, you just need to sum those angles to the latitude and longitude of your position.

With your data:

&gt; # Earth radius in chains
&gt; (r &lt;- 3958.8*80)
[1] 316704
&gt; 
&gt; #bearing, latitude and longitude in radians
&gt; (br &lt;- -52/180*pi)
[1] -0.907571211
&gt; (latr &lt;- 37/180*pi)
[1] 0.6457718232
&gt; (lonr &lt;- -91/180*pi)
[1] -1.588249619
&gt; 
&gt; #1 Distance to North and East in chains
&gt; (east &lt;- sin(br)*2.67)
[1] -2.103988712
&gt; (north &lt;- cos(br)*2.67)
[1] 1.643816139
&gt; 
&gt; #One degree in chains
&gt; (deglat &lt;- r*pi/180)
[1] 5527.527554
&gt; (deglon &lt;- r*pi/180*cos(latr))
[1] 4414.479788
&gt; 
&gt; #2 distance in degrees
&gt; (dlat &lt;- north/deglat)
[1] 0.0002973872356
&gt; (dlon &lt;- east/deglon)
[1] -0.000476610793
&gt; 
&gt; #3 arriving point (GPS coordinates)
&gt; (lat &lt;- 37+dlat)
[1] 37.00029739
&gt; (lon &lt;- -91+dlon)
[1] -91.00047661