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?
On
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 $3\angle A + 2\angle B = 180^\circ$
$$\frac{10}{\cos \left(\frac{45.6^\circ}{2}\right)} = \frac{BC}{\sin 45.6^\circ}$$
$BC =7.75$
Hint:
You can use the Law of Sines and the Law of Cosines