Question 919179
Having the measures of one angle and the opposite side, you can use law of sines to try to solve the triangle.
In this case, you can use A, a and b to find B, C and c).
Law of sines says that {{{sin(A)/a=sin(B)/b}}} , so
{{{sin(30^o)/10=sin(B)/9}}}-->{{{0.5/10=sin(B)/9}}}-->{{{9*0.5/10=sin(B)}}}-->{{{sin(B)=0.45}}}
That means that {{{highlight(B=26.74^o)}}} (rounded).
That is the only possible measure for {{{B}}} for the only possible triangle with the given measurements.
{{{sin(180^o-26.74^o)=sin(153.26^o)=0.45}}} too,
but you could not have angles measuring {{{30^o}}} and {{{153.26^o}}} in the same triangle,
because {{{30^o+153.26^o=183.26^o>180^o}}}.
So there is only one option, one triangle,
just one {{{B}}} and one {{{C}}} , no {{{B1}}} or {{{C1}}} .
{{{C=180^o-(A+B)}}}-->{{{C=180^o-(30^o+26.74^o)}}}-->{{{C=180^o-56.74^o}}}-->{{{highlight(C=123.26^o)}}} (rounded).
Law of sines also says {{{sin(A)/a=sin(C)/c}}}<-->{{{a/sin(A)=c/sin(C)}}} , so
{{{10/0.5=c/sin(123.26^o)}}}-->{{{20=c/0.8362}}}(rounded)-->{{{c=20*0.8362}}}(rounded)-->{{{highlight(c=16.7)}}} (rounded).
 
When you draw the {{{30^o}}} angle, you can locate
point A, at the vertex of the angle,
ray AB on one side of the angle,
and point C, at distance {{{b=9}}} from A on the other side of the angle (side AC).
You know that point B is at distance {{{a=10}}} from point C,
and that it is on ray AB,
so you just draw an arc with radius {{{a=10}}} centered at point C.
The point where that arc intersects ray AB is point B.
angle{{{drawing(400,150,-1,19,-2,5.5,
arrow(0,0,18.9,0),line(0,0,7.79,4.5),locate(7.7,5.2,C),
green(arc(7.79,4.5,20,20,-10,40)),locate(12.5,5,green(10)),
green(arrow(13.5,4.5,17.79,4.5)),green(arrow(12.5,4.5,7.79,4.5)),
locate(-0.4,0,A),locate(1.1,1.2,red(30^o)),
red(arc(0,0,5,5,-30,0)),locate(4,2.5,b=9)
)}}}