Question 398510
<pre><font face = "batangche" color = "indigo" size = 4><b>
We draw the triangle:

{{{drawing(800,240,-2,7,-.5,2.2,  triangle(0,0,-1,sqrt(3),6,0),
locate(0,.2,"120°"), locate(0,0,C), locate(3,0,6), locate(6,0,B),
locate(-1.1,1.9,A), locate(-.65,.9,2),
red(arc(0,0,1,-1,0,120)) 
)}}}

We extend BC, and draw a line segment from A perpendicular to the 
extension of BC, intersecting it at D:

{{{drawing(800,240,-2,7,-.5,2.2,  triangle(0,0,-1,sqrt(3),6,0),
locate(0,.2,"120°"), locate(0,0,C), locate(3,0,6), locate(6,0,B),
locate(-1.1,1.9,A), locate(-.65,.9,2), rectangle(-1,0,-.9,.1),
green(line(0,0,-1,0),line(-1,0,-1,sqrt(3))), locate(-1,0,D)
red(arc(0,0,1,-1,0,120)) 

)}}}

Angle ACD is 60° because it is supplementary to angle ACB which is 120°,
and 180°- 120° = 60°

{{{drawing(800,240,-2,7,-.5,2.2,  triangle(0,0,-1,sqrt(3),6,0),
locate(0,.2,"120°"), locate(0,0,C), locate(3,0,6), locate(6,0,B),
locate(-1.1,1.9,A), locate(-.65,.9,2), rectangle(-1,0,-.9,.1),
green(line(0,0,-1,0),line(-1,0,-1,sqrt(3))), locate(-1,0,D)
red(arc(0,0,1,-1,0,120)), green(locate(-.4,.2,"60°"), arc(0,0,1,-1,120,180)) 

)}}}
 
Triangle ADC is a "30°60°90°" right triangle, and since its          _
hypotenuse AC is 2, its lower leg CD is 1, and its vertical leg AD is &#8730;3.
BD = BC + CD = 6 + 1 = 7 

{{{drawing(800,240,-2,7,-.5,2.2,  triangle(0,0,-1,sqrt(3),6,0),
locate(0,.2,"120°"), locate(0,0,C), locate(3,0,6), locate(6,0,B),
locate(-1.1,1.9,A), locate(-.65,.9,2), rectangle(-1,0,-.9,.1),
green(line(0,0,-1,0),line(-1,0,-1,sqrt(3))), locate(-1,0,D)
red(arc(0,0,1,-1,0,120)), green(locate(-.4,.2,"60°"), arc(0,0,1,-1,120,180),
locate(-.5,0,1),locate(-1.3,.9,sqrt(3))) 

)}}}

Triangle ADB is a right triangle, therefore

tan(B) = {{{AD/BD}}} = {{{sqrt(3)/7}}}

Use inverse tangent function to get

Angle B = 13.89788625° closest to 14°.

Edwin</pre>