Question 959109
THERE ARE TWO special triangles in trigonometry. One is the {{{30}}}°-{{{60}}}°-{{{90}}}° triangle. The other is the {{{isosceles}}}{{{ right}}}{{{ triangle}}}. They are special because, with simple geometry, we can know the ratios of their sides.


so, what we need to do first is find the sides length using given points and distance formula

(1,1), (4,1), and (3,4)

the distance between (1,1)and (4,1) 

{{{d=sqrt((4-1)^2+(1-1)^2)}}}

{{{d=sqrt(3^2+0^2)}}}

{{{d=sqrt(3^2)}}}

{{{d=3}}}


the distance between (1,1)and (3,4) 

{{{d=sqrt((3-1)^2+(4-1)^2)}}}

{{{d=sqrt(2^2+3^2)}}}

{{{d=sqrt(13)}}}-exactly

{{{d=3.6}}}-approximately


the distance between (4,1) and (3,4)

{{{d=sqrt((4-3)^2+(1-4)^2)}}}

{{{d=sqrt(1^2+(-3)^2)}}}

{{{d=sqrt(10)}}}-exactly

{{{d=3.2}}}-approximately

so, the shortest side is  {{{3}}} units long, then the longer side is  {{{sqrt(10)}}} and the longest side is  {{{sqrt(13)}}} 


now, check if these are sides of a {{{30}}}°-{{{60}}}°-{{{90}}}° triangle

first, use Pythagorean theorem because triangle has an {{{90}}}° angle

{{{(sqrt(13))^2=(sqrt(10))^2+3^2}}}

{{{13=10+9}}}

{{{13<>19}}}=>your triangle is not a {{{30}}}°-{{{60}}}°-{{{90}}}° triangle

other way is to use a theorem: 

 In a {{{30}}}°-{{{60}}}°-{{{90}}}° triangle the sides are in the ratio {{{1 : 2 : sqrt( 3)}}}.

now, the sides of this triangle are in ratio {{{3:sqrt(10):sqrt(13)}}} 

compared to {{{1 : 2 : sqrt( 3)}}}, we can see that  these ratios are not equivalent; so, your triangle is not a 
{{{30}}}°-{{{60}}}°-{{{90}}}° triangle 


now, see it on a graph:

{{{drawing( 600, 600, -5, 5, -5, 5,

circle(1,1,.03),circle(4,1,.03),circle(3,4,.03),

locate(1,1,A),locate(4,1,B),locate(3,4.3,C),

line(1,1,4,1),line(1,1,3,4),line(3,4,4,1),
 graph( 600, 600, -5, 5, -5, 5, 0)) }}}