Find the number of distinct triangles with the measurements a=1 b=2 and A=31 degrees

we won't have any triangle with these measures.

law of cosines.

a²=b²+c²-2bc (cosФ)

dа ta:

a=1

b=2

A=31º

Therefore:

1²=2²+c²-4c (cos 31º)

c² - (4cos 31º) c+3=0

4cos 31º≈3.43

We have to solve this equation, and find out the number of solutions:

c=[4cos 31º⁺₋√ (11.756-12) ]/2=

c = (4cos 31º⁺₋√ (-0.244) / 2

Because we have the square root of a negative number, we don't have any possible solutions, that is to say, there are not real solutions for this equation.

Answer: We don't have any triangle with these measures.