Find side BC of a triangle given AB, AC, and a relation between $\angle A$ and $\angle B$

A question from my class:

In triangle $ABC$, $3\angle A+2\angle B=180$ and $AB=10, AC=4$. So question is, what all can we comment on side $BC$. Can we find its exact length?

I have a crude solution involving trigonometry and a equation, but it's too large. So, can anybody help me?

• If it makes sense, then it's fine. Maybe you are searching for a more elegant way, then I suggest you to report your solution and specify your new requirement – mattecapu Aug 2 '14 at 22:07

Hint:

You can use the Law of Sines and the Law of Cosines

• I did, but it became a big equation. so I had to quit that method – Dinesh Aug 2 '14 at 17:33
• You can express $\angle C$ and $\angle B$ in terms of $\angle A$. Using the Law of sines and the formula of sine of sum of angles you get to: $$\frac{10}{\cos (A/2)} = \frac{4}{\sin (3A/2)} = \frac{BC}{\sin A}$$ – Darth Geek Aug 2 '14 at 17:38

Let $\frac{A}{2} = \beta$

$10(4\cos^3\beta - 3\cos \beta)= 4\cos \beta$

Since $\cos\beta$ is not equal to zero (if it were equal to zero, then A would be $180^\circ$ and the triangle would not exist),

$40\cos^2\beta=34$

$\cos^2\beta=\frac{17}{20}$, $\sin^2\beta=\frac{3}{20}$

$\cos A= \cos 2 \beta = \cos^2\beta - \sin^2\beta = 0.7$

$\angle A = 45.6^\circ$

$\angle B = 21.6^\circ$

It is not necessary to know the value of $\angle B$ to determine the value of $BC$, but it helps to verify that $3\angle A + 2\angle B = 180^\circ$

$$\frac{10}{\cos \left(\frac{45.6^\circ}{2}\right)} = \frac{BC}{\sin 45.6^\circ}$$

$BC =7.75$