The triangle is a right triangle with the right angle being at B.
We can use the law of sines to find the other angles and lengths.
law of sines says:
a/sin(A) = b/sin(B) = c/sin(C)
we know a and b and we know A, so we can use that formula to find B
a/sin(A) = b/sin(B) becomes 1/sin(A) = 2/sin(B) which becomes 1/sin(30) = 2/sin(B).
we cross multiply to get sin(B) = 2 * sin(30) which is equal to 2 * .5 which is equal to 1.
we know that sin(B) = 1 which means that B = 90 degrees.
Since we know that angle B is 90 degrees and angle A is 30 degrees and we know that the sum of the angles of a triangle is equal to 180 degrees, then angle C has to be equal to 60 degrees.
We can use the law of sines again to find the length of c.
c/sin(C) = a/sin(A) which becomes c/sin(60) = 1/sin(30).
Cross multiply to get c*sin(30) = sin(60)
divide both sides of the equation by sin(30) to get c = sin(60)/sin(30).
this makes c = 1.732050808 which is equal to square root of 3.
I know that because when I square 1.732050808 I get 3.
I also looked at this graphically to see if I could determine how many solutions are possible from the graph.
The following graph shows the result of this investigation.
When x = 2 and y = 0, we draw a circle around that point with a radius of 1.
We also draw a line with the equation of y = tan(30)*x.
We use the x -axis as the third line.
The intersection of the circle with the line tells us how many points we can use for our triangle.
It looks like we have only 1 point of intersection between the circle and the line with the slope of tan(30).
It also looks like the line is actually tangent to the circle at that point.
this is confirmed by previous analysis that shows that angle to be 90 degrees, which it would be if the line were tangent to the circle at that point.
The angle I am talking about is the angle of the line with the intersection of of the radius at that point.
The following picture shows what I am talking about.
Triangle is labeled ABC.
Intersection with the circle is at B.
The line AB is perpendicular to the line BC at point B.
