# Ratio of angles in a triangle, given lengths of triangle's sides. [closed]

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 $\,\dfrac{\angle ACB}{\angle BAC}\,$?

## closed as off-topic by José Carlos Santos, Abcd, Paul Frost, A. Pongrácz, GibbsDec 30 '18 at 22:01

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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 $$\dfrac 6{\sin(\angle ACB)} = \dfrac 4{\sin(\angle BAC)} \iff \dfrac {6}{4} = \dfrac{\sin(\angle ACB)}{\sin(\angle BAC)} = \frac 32$$

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

• But how do you use the ratio of sines of angles to compute the ratio of angles? – Adam Nov 12 '13 at 14:10
• To determine the ratio of the angles directly, use the law of cosines: you have the lengths of each side (equivalent to a, b, c in the formula above.) Use them to calculate the respective angle, for each angle, by using, e.g., for $\gamma$ above, that would be $$\gamma = \cos^{-1}\left(\dfrac{a^2 + b^2 - c^2}{2ab}\right)$$ – Namaste Nov 12 '13 at 14:16
• $$\angle ACB = \cos^{-1}\left(\frac{4^2 + 5^2 - 6^2}{2\cdot 4 \cdot 5}\right)$$ $$\angle BAC = \cos^{-1}\left(\frac{5^2 + 6^2 - 4^2}{2\cdot 5\cdot 6}\right)$$ $$\frac{\angle ACB}{\angle BAC} = \frac{\cos^{-1}\left(\frac{4^2 + 5^2 - 6^2}{2\cdot 4 \cdot 5}\right)}{\cos^{-1}\left(\frac{5^2 + 6^2 - 4^2}{2\cdot 5\cdot 6}\right)}$$ – Namaste Nov 12 '13 at 14:21
• $\angle ACB = \cos^{-1}\frac18$ and $\angle BAC = \cos^{-1}\frac34$. So $\sin \angle BAC = \sqrt{1 - \frac9{16}} = \frac{\sqrt7}2$. By the double angle formula, $\cos (2\angle BAC) = (\frac34)^2 - (\frac{\sqrt7}2)^2 = \frac18 = \cos \angle ACB$. – NovaDenizen Jan 22 '14 at 15:40