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

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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?

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2 Answers

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Hint:

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

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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 $3\angle A + 2\angle B = 180^\circ$

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

$BC =7.75$

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