Ratio of angles in a triangle, given lengths of triangle's sides.

Question

If I have a triangle $\,\triangle ABC,\,$ with sides of lengths $\,AB=6, \;BC=4, \;CA=5,\,$

then what can I know about the ratio of $\,\dfrac{\angle ACB}{\angle BAC}\,$?

Answer 1

You can use the Law of Sines to find the ratio of the sines of your two angles:

$$\frac{a}{\sin A} \,=\, \frac{b}{\sin B} \,=\, \frac{c}{\sin C} \,=\, D \! $$

In your case, you'd have $$\dfrac 6{\sin(\angle ACB)} = \dfrac 4{\sin(\angle BAC)} \iff \dfrac {6}{4} = \dfrac{\sin(\angle ACB)}{\sin(\angle BAC)} = \frac 32$$

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Alternatively, to compute the measures of your angles directly, use the Law of Cosines. $$c^2 = a^2 + b^2 - 2ab\cos\gamma\,\iff \cos \gamma = \dfrac{a^2 + b^2 - c^2}{2ab}$$