#### Dj0rdjev1c

##### New member

- Joined
- Apr 7, 2019

- Messages
- 2

- Thread starter Dj0rdjev1c
- Start date

- Joined
- Apr 7, 2019

- Messages
- 2

- Joined
- Nov 24, 2012

- Messages
- 1,312

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?

- Joined
- Apr 7, 2019

- Messages
- 2

Ty for help