Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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,

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

Similarly,

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

Therefore,

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

Also,

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

Substituting:

\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?

share|improve this question

1 Answer 1

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}

share|improve this answer
    
Thanks, could you add why this result is equivalent to mine? –  timvermeulen 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
1  
I got it, and edited your answer. –  timvermeulen Jun 26 '13 at 17:48
    
@timvermeulen I've edited in your proof. Very nice. –  Adriano Jun 26 '13 at 17:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.