# Finding the unknown side of a triangle

#### Dj0rdjev1c

##### New member
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

##### Super Moderator
Staff member
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

##### Super Moderator
Staff member
$$\displaystyle 3a^2=2a^2+10$$
$$\displaystyle a=\sqrt{10}$$