Finding the unknown side of a triangle

Dj0rdjev1c

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Text: In triangle ABC, angle A is 2 times bigger than angle B, AC=2cm, AB=3cm, find BC
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This is all that i can see on this triangle, can someone help me please?
 

MarkFL

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Hello, and welcome to FMH! :)

Suppose we set:

\(\displaystyle \angle A=\alpha=2x\)

Then:

\(\displaystyle \angle B=\beta=x\)

And thus:

\(\displaystyle \angle C=\gamma\implies \gamma=\pi-3x\)

Let:

\(\displaystyle \overline{BC}=a\)

Using the Law of Sines, we may then state:

\(\displaystyle \frac{a}{\sin(2x)}=\frac{2}{\sin(x)}\implies a=4\cos(x)\)

Using the Law of Cosines, we may state:

\(\displaystyle 2^2=3^2+a^2-6a\cos(x)\implies \cos(x)=\frac{a^2+5}{6a}\)

And so there results:

\(\displaystyle a=4\left(\frac{a^2+5}{6a}\right)\)

Can you proceed?
 

MarkFL

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To follow up...

\(\displaystyle 3a^2=2a^2+10\)

\(\displaystyle a=\sqrt{10}\)
 

Dj0rdjev1c

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Ty for help
 
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