Question 1035792
I have to solve a triangle. Given 2 sides and 1 angle from 
it. I gotta determine if it's based on 1 triangle, 
2 triangles, or no triangles. a= 19, b= 10, B= 10 degrees.
<pre><b>
To get a better understanding, let's sketch the geometry of the 
situation. I start by drawing an arbitrarily long line labeling 
the endpoint as the vertex of the given angle, in this case B.

{{{drawing(800,5200/19,-5,33,-5,8,
locate(0,0,B),

line(0,0,40,0) )}}}

At point B we draw the angle B = 10° by drawing side 
BC = a = 19:

{{{drawing(800,5200/19,-5,33,-8,5,
locate(0,0,B),locate(18.7,4.2,C),
triangle(0,0,0,0,18.71134731,3.299315376),
locate(7.6,2.3,a=19),locate(3.7,.76,"10°"),
red(arc(0,0,12,-12,0,10)),
line(0,0,40,0) )}}}

Next we take a compass and open it to the length
b = CA = 10.  Put the sharp point on C and swing 
an arc that cuts the horizontal line that we drew
first in two places:

{{{drawing(800,5200/19,-5,33,-8,5,
locate(0,0,B),locate(18.7,4.2,C),
triangle(0,0,0,0,18.71134731,3.299315376),
locate(7.6,2.3,a=19),locate(3.7,.76,"10°"),
red(arc(0,0,12,-12,0,10),arc(18.71134731,3.299315376,20,-20,197,348)),
line(0,0,40,0) )}}}

So it appears in this case there are two choices for point A.  
Either this way with the green line:

{{{drawing(800,5200/19,-5,33,-8,5,
locate(0,0,B),locate(18.7,4.2,C),green(line(18.71134731,3.299315376,9.271298679,0),locate(9,0,A),locate(14.3,1.9,b=10)),
triangle(0,0,0,0,18.71134731,3.299315376),
locate(7.6,2.3,a=19),locate(3.7,.76,"10°"),
red(arc(0,0,12,-12,0,10),arc(18.71134731,3.299315376,20,-20,197,348)),
line(0,0,40,0) )}}}
 
or this way with the blue line:

{{{drawing(800,5200/19,-5,33,-8,5,
locate(0,0,B),locate(18.7,4.2,C),blue(line(18.71134731,3.299315376,28.15139594,0),locate(28.2,0,A),locate(23,2.6,b=10)),
triangle(0,0,0,0,18.71134731,3.299315376),
locate(7.6,2.3,a=19),locate(3.7,.76,"10°"),
red(arc(0,0,12,-12,0,10),arc(18.71134731,3.299315376,20,-20,197,348)),
line(0,0,40,0) )}}}

Now let's do the calculating using the law of sines:

{{{matrix(1,5,
a/sin(A),""="",b/sin(B),""="",c/sin(C))}}}

substitute a=19, b=10, B=10°

{{{matrix(1,5,
19/sin(A),""="",10/sin("10°"),""="",c/sin(C))}}}

We use only the first and second parts:

{{{matrix(1,3,
19/sin(A),""="",10/sin("10°"))}}}

Cross-multiply

{{{matrix(1,3,
10sin(A),""="",19sin("10°"))}}}

Divide both sides by 10

{{{matrix(1,3,
sin(A),""="",19sin("10°")/10)}}}

Do the calculation:

{{{matrix(1,3,
sin(A),""="",0.3299315376)}}}

[If this had come out greater than 1 there would have 
been 0 solutions.  If this had come out equal to 1 
there would have been 1 solution, and it would have 
been a right triangle with hypotenuse b.]

Since it came out less than 1, we can tell there is
either 1 or 2 solutions.

First solution:
Next we calculate the first quadrant (acute) solution
for angle A using inverse sine feature on the calculator
and get angle A = 19.26462015°.  Then we can calculate the
third angle C by adding 19.26462015° and 10° and subtracting
from 180°, getting angle C = 150.7353799°.

Now we use the law of sines again:

{{{matrix(1,5,
19/sin(A),""="",10/sin("10°"),""="",c/sin(C))}}}

This time we use only the second and third parts,
and substitute 150.7353799° for angle C 

{{{matrix(1,3,
10/sin("10°"),""="",c/sin("150.7353799°"))}}}

Cross-multiply:

{{{matrix(1,3,
c*sin("10°"),""="",10*sin("150.7353799°"))}}}

Divide both sides by sin(10°)

{{{matrix(1,3,
c,""="",10*sin("150.7353799°")/sin("10°"))}}}

{{{matrix(1,3,
c,""="",28.15139588)}}}

-----------------

Next we see if there is a second solution.
We calculate the second quadrant (obtuse) solution
for angle A by finding the complement of 19.26462015°,
by subtracting it from 180°, getting 160.7353798°.
Then we can calculate the third angle C by adding 
160.7353798° and 10° and subtracting
from 180°, getting angle C = 9.2646202°.

[If this had come out negative there would have 
been only the 1 solution.]

So we have a second solution, and we use the law of
sines to find c just as we did to find the first
solution:

{{{matrix(1,3,
10/sin("10°"),""="",c/sin("9.2646202°"))}}}

Cross-multiply:

{{{matrix(1,3,
c*sin("10°"),""="",10*sin("9.2646202°"))}}}

Divide both sides by sin(10°)

{{{matrix(1,3,
c,""="",10*sin("9.2646202°")/sin("10°"))}}}

{{{matrix(1,3,
c,""="",9.271298727)}}}

Edwin</pre></b>