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I wondered what the ratios between the sides of a triangle is, when the angles are known. So basically:

$\triangle ABC$ has angles $\alpha, \beta \text{ and } \gamma$. Find $\frac{\lvert AB \rvert}{\lvert AC \rvert}$.

A line through $C$ perpendicular to $AB$ intersects $AB$ at point $D$. By definition,

$$ \tan \alpha = \frac{\lvert CD \rvert}{\lvert AD \rvert} \implies \lvert AD \rvert = \lvert CD \rvert \cdot \cot \alpha $$


$$ \lvert BD \rvert = \lvert CD \rvert \cdot \cot \beta $$


$$ \lvert AB \rvert = \lvert AD \rvert + \lvert BD \rvert = \lvert CD \rvert \cdot \left(\cot \alpha + \cot \beta\right) $$


$$ \sin \alpha = \frac{\lvert CD \rvert}{\lvert AC \rvert} \implies \lvert CD \rvert = \sin\alpha \cdot \lvert AC \rvert $$


\begin{align*} \lvert AB \rvert &= \sin\alpha \cdot \lvert AC \rvert \cdot \left(\cot \alpha + \cot \beta\right)\\ &= \lvert AC \rvert \cdot \left( \sin\alpha \cdot \cot\alpha + \sin\alpha \cdot \cot\beta \right)\\ &= \lvert AC \rvert \cdot \left( \cos\alpha + \frac{\sin\alpha}{\tan\beta} \right) \end{align*}

And thus,

$$ \frac{\lvert AB \rvert}{\lvert AC \rvert} = \cos\alpha + \frac{\sin\alpha}{\tan\beta} $$

This seems like such a strange result, which I have never seen before. Is it correct? If so, is it usually expressed differently?

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By Sine Law, we have: $$ \dfrac{|AB|}{\sin\gamma}=\dfrac{|AC|}{\sin\beta} \iff \boxed{\dfrac{|AB|}{|AC|}=\dfrac{\sin \gamma}{\sin\beta}} $$

To see why this works, construct a line through $A$ perpendicular to $BC$. This new line will intersect $BC$ at a point; call it $E$. Then by definition, we have: $$ \sin \beta = \dfrac{|AE|}{|AB|} \qquad \text{and} \qquad \sin \gamma = \dfrac{|AE|}{|AC|} $$ Hence, we have: $$ |AB|\sin\beta=|AE|=|AC|\sin\gamma $$ which can be used to derive both Sine Law as well as our desired result.

OP Edit: Indeed, $\cos\alpha + \dfrac{\sin\alpha}{\tan\beta} = \dfrac{\sin\gamma}{\sin\beta}$:

\begin{align*} \cos\alpha + \frac{\sin\alpha}{\tan\beta} &= \frac{\cos\alpha \cdot \sin\beta}{\sin\beta} + \frac{\sin\alpha \cdot \cos\beta}{\sin\beta}\\ &= \frac{\cos\alpha \cdot \sin\beta + \sin\alpha \cdot \cos\beta}{\sin\beta}\\ &= \frac{\sin\left(\alpha+\beta\right)}{\sin\beta}\\ &= \frac{\sin\left(\pi - \left(\alpha+\beta\right)\right)}{\sin\beta}\\ &= \frac{\sin\gamma}{\sin\beta} \end{align*}

Using two theorems:

\begin{equation} \sin\alpha \cdot \cos\beta + \cos\alpha \cdot \sin\beta = \sin\left(\alpha + \beta\right) \end{equation}

\begin{equation} \sin\theta = \sin\left(\pi-\theta\right) \end{equation}

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Thanks, could you add why this result is equivalent to mine? – Tim Vermeulen Jun 26 '13 at 17:20
@timvermeulen Your work looks correct. I can't think of a direct way to prove their equivalence off the top of my head; it might involve using the fact that $\alpha+\beta+\gamma=\pi$ and using some kind of angle sum/difference identity, which I'm not very familiar with. – Adriano Jun 26 '13 at 17:33
I got it, and edited your answer. – Tim Vermeulen Jun 26 '13 at 17:48
@timvermeulen I've edited in your proof. Very nice. – Adriano Jun 26 '13 at 17:59

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