Is a triangle possible with respect to side, angle and proportional sides?

Question

Mathematics

Hello.
I'm in a clinch. I don't even know which way to dig.
There are three segments: a, b, c.
The segments a and b form a known angle A between themselves.
The length of the segment a is also known.
And it is known that b/c = k. This k is also known.
So. Under what condition will it be possible to form a triangle from these segments?
Thank you!

Answer 1

A triangle can be made up of any three segments as long as the triangle inequality a+b > c, a+c > b, b+c > a is satisfied. it is enough to check only one of them - the maximum number must be less than the sum of the rest. But if it is not obvious which of them is the maximum, then you can check all three.
So only A and k are given, right?
To somehow connect the angle with the sides, you need to use the cosine theorem :
c^2 = a^2+b^2-2*a*b*cosA.
substituting judgment b=ck, we get
c^2 = a^2+k^2c^2-2*k*a*c*cosA
This is a quadratic equation relating C and A. You can solve it for a (c is a parameter) and you get a expressed in terms of c.
Well, in the end, you need to check that a + b > c, a + c > b, b + c > a, substituting the found formulas for b and a through c. In these inequalities there will be known cosA, sinA, k and unknown c. Since the angle and aspect ratio of a triangle can be scaled, the inequality must hold for all c. But in fact, in these inequalities, all c can be stupidly reduced. Some of them can be immediately thrown out if we consider 2 cases k>1 and k<=1. Well, how to solve them further - think for yourself.