Question

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

Answer 1

we won´t have any triangle with these measures.

law of cosines.
a²=b²+c²-2bc(cosФ)

Data:
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.

Answer 2

No triangle possible.

Explanation:

Let in triangle ABC,

BC = a = 1 unit,

AC = b = 2 unit,

m∠A = 31°,

By the law of sine,

`(sin B)/(b)=(sin A)/(a)`

`\implies sin B=(2sin 31^(\circ))/(1)=2sin 31^(\circ)=1.0301`

⇒ m∠B= undefined

( ∵ the value of sin Ф lies from 0 to 1 where Ф is an angle )

An angle of a triangle can not be undefined.

∴ Triangle ABC is not possible,

That is, there is no possible triangle with the given measures.