Finding the unknown side of a triangle

[Dj0rdjev1c]

Text: In triangle ABC, angle A is 2 times bigger than angle B, AC=2cm, AB=3cm, find BC

This is all that i can see on this triangle, can someone help me please?

 

[MarkFL]

Hello, and welcome to FMH!

Suppose we set:

[MATH]\angle A=\alpha=2x[/MATH]
Then:

[MATH]\angle B=\beta=x[/MATH]
And thus:

[MATH]\angle C=\gamma\implies \gamma=\pi-3x[/MATH]
Let:

[MATH]\overline{BC}=a[/MATH]
Using the Law of Sines, we may then state:

[MATH]\frac{a}{\sin(2x)}=\frac{2}{\sin(x)}\implies a=4\cos(x)[/MATH]
Using the Law of Cosines, we may state:

[MATH]2^2=3^2+a^2-6a\cos(x)\implies \cos(x)=\frac{a^2+5}{6a}[/MATH]
And so there results:

[MATH]a=4\left(\frac{a^2+5}{6a}\right)[/MATH]
Can you proceed?

 

[MarkFL]

To follow up...

[MATH]3a^2=2a^2+10[/MATH]
[MATH]a=\sqrt{10}[/MATH]

 

[Dj0rdjev1c]

Ty for help