Question 205114
<font size = 8 color = "red"><b>Edwin's solution. Stanbon is wrong to say angle A could be 150°. It can only be 30°.  Here's why:</b></font>

{{{drawing(400,200,-2,3,-.4,2.1, locate(0,0,B),locate(2,0,C),
locate(-1,1.9,A),locate(0,.25,"120°"), locate(1,0,x),
 black(arc(0,0,1,-1,0,120)), locate(.4,1.1,x*sqrt(3)),
triangle(0,0,2,0,-1,sqrt(3)) )}}}
<pre><font size = 4 color = "indigo"><b>
By the law of sines:

{{{BC/sin(A)=AC/sin(B)}}}


{{{x/sin(A)=(x*sqrt(3))/sin("120°")}}} 

Cross-multiply:

{{{x*sqrt(3)sin(A)=x*sin("120°")}}}

Divide both sides by {{{x*sqrt(3)}}}

{{{(x*sqrt(3)sin(A))/(x*sqrt(3))=(x*sin("120°"))/(x*sqrt(3))}}}

{{{(cross(x*sqrt(3))sin(A))/cross(x*sqrt(3))=(cross(x)*sin("120°"))/(cross(x)*sqrt(3))}}}

{{{sin(A)=sin("120°")/sqrt(3)}}}

We know that {{{sin("120°")=sqrt(3)/2}}}, so substituting:

{{{sin(A)=(sqrt(3)/2)/sqrt(3)}}}

Write as a division:

{{{sin(A)=(sqrt(3)/2)}}}{{{"÷"sqrt(3)}}}

Invert and multiply

{{{sin(A)=(sqrt(3)/2)}}}{{{"*"(1/sqrt(3))}}}

{{{sin(A)=(cross(sqrt(3))/2)}}}{{{"*"(1/cross(sqrt(3)))}}}

{{{sin(A)=1/2}}}

There are two angles between 0° and 180° which
have sine {{{1/2}}}, they are 30° and 150°, but

angle A cannot be 150°, because angle B is 120°,
and that would make two angles in the same
triangle with sum more than 180° which is 
impossible.  So angle A can only be 30°.

Edwin</pre>