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If in a $\displaystyle\bigtriangleup$ ABC, $\displaystyle\cot \biggl( \frac{A}{2} \biggr) = \frac{b+c}{a} $, then $\displaystyle\bigtriangleup$ ABC is of which type ?

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This is tough. I got as far as establishing the relation $a(b+c)=bc(1+\cos\;A)\tan\;A$, but can't figure out the required inequalities. – J. M. Nov 16 '10 at 10:31
@J.M: In my module they have explained it to be as right-angled triangle, but as of now I am yet to understand that explanation. – Quixotic Nov 16 '10 at 10:48
@J.M.: unfold one side (say, "b") onto the other to get a triangle with sides (a) and (b+c) and an angle (A/2). If the ratio of sides is the same as for a right triangle, that is that. – T.. Nov 17 '10 at 0:10
I got the idea after seeing the excellent answers already given (and I've voted all of them up), but thanks for that, @T. :) (I said inequality in that first comment since I was hoping to turn the original relation into something that looked like the law of cosines, but got mired in the algebra) – J. M. Nov 17 '10 at 0:59
up vote 8 down vote accepted

So by Law of Sines we have $$ \frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} =k (\text{say})$$

From this your equation becomes, $$\frac{\cos\frac{A}{2}}{\sin\frac{A}{2}} = \frac{k(\sin{B} + \sin{C})}{k \sin{A}} = \frac{\sin{B}+\sin{C}}{\sin{A}} = \frac{\cos\frac{A}{2} \cos\frac{B-C}{2}}{\sin\frac{A}{2} \cos\frac{A}{2}}$$ From this you get $$\frac{\pi}{2} - \Bigl(\frac{B+C}{2}\Bigr)= \pm \frac{B-C}{2}$$ and you can do it then.

This gives $B=\frac{\pi}{2}$ or $C =\frac{\pi}{2}$, which means the triangle is right-angled. Hurray!

NOTE: The keen idea when you see problems of this type is to use the Sine rule. Do you think that this problem would have been difficult, if you had applied the sine rule as i did in the first step. Very often in Mathematics starting the solution is the difficult part. Once you figure that out, then the solutions simply flows.

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@Chandru: You can't conclude that $B$ is definitively the right angle when there's nothing in the problem that distinguishes $B$ from $C$ (or $b$ from $c$). You should point out any extra facts you're assuming (say, that $B \ge C$), and be clear when this fact comes into play. (Or, leave an appropriate sign ambiguity in your equations and conclude that "either $B$ or $C$ is $\pi/2$".) Nevertheless, I like this approach; it's more trig than algebra, which always seems more satisfying to me. +1. – Blue Nov 16 '10 at 12:19
your first equation .. is it Napier's analogy or sine rule ? –However,+1 and accepted since very precise explanation and a good tip in the end :) – Quixotic Nov 16 '10 at 13:29
@Debajan: I learn't this in high school as Napier's Analogy. – anonymous Nov 16 '10 at 14:59
@Day Late Don: Fixed. – Aryabhata Nov 16 '10 at 18:45
@Moron: Help ke liye shukriya. – anonymous Nov 16 '10 at 20:56

The cosine formula states

$$a^2 = b^2+ c^2 – 2bc \cos A$$

and since

$$\cos A = \frac{ \cot^2 (A/2) – 1}{ \cot^2(A/2)+1}$$

we have

$$a^2 = b^2 + c^2 -2bc \frac{ (b+c)^2 – a^2}{(b+c)^2 + a^2},$$

which reduces to $a^4 = (b^2 – c^2)^2.$

And so taking the square root $a^2 =b^2 – c ^2,$ where $b$ is the larger side.

Hence $b^2 = c^2+a^2$ and so we have a right-angled triangle.

And so it's the angle opposite the larger of $b$ and $c$ which is the right-angle.

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Begin with the Law of Cosines to compute $\cot^2\frac{A}{2}$ in terms of the lengths of the sides of the triangle:

$$\cos{A} = \frac{b^2+c^2-a^2}{2bc}$$ $$\Rightarrow \begin{cases} \cos^2{\frac{A}{2}}=\frac{1+\cos{A}}{2}=\frac{2bc+b^2+c^2-a^2}{4bc} = \frac{(b+c)^2-a^2}{4bc} \\ \sin^2{\frac{A}{2}}=\frac{1-\cos{A}}{2}=\frac{2bc-(b^2+c^2-a^2)}{4bc} = \frac{a^2-(b-c)^2}{4bc} \end{cases}$$ $$\Rightarrow \cot^2\frac{A}{2}=\frac{\cos^2{\frac{A}{2}}}{\sin^2{\frac{A}{2}}}=\frac{(b+c)^2-a^2}{a^2-(b-c)^2}$$

Now, work in the condition from the problem, and let Algebra bring you home ...

$$\begin{eqnarray} \frac{(b+c)^2}{a^2} &=& \frac{(b+c)^2-a^2}{a^2-(b-c)^2} \\ (b+c)^2\left(a^2-(b-c)^2\right)&=& a^2\left((b+c)^2-a^2\right) \\ a^2(b+c)^2-(b+c)^2(b-c)^2 &=& a^2 (b+c)^2-a^4 \\ a^4-(b+c)^2(b-c)^2 &=& 0 \\ a^4-(b^2-c^2)^2 &=& 0 \\ \left(a^2+(b^2-c^2)\right)\left(a^2-(b^2-c^2)\right) &=& 0 \\ \end{eqnarray}$$ $$\Rightarrow \hspace{0.25in} a^2+b^2=c^2 \hspace{0.25in}\text{or}\hspace{0.25in} a^2+c^2=b^2$$

Nice little problem. I suspect there's a more-direct path to the solution, though. (Edit I see a couple such paths had been posted when I was working on my answer! :)

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OK, first rewrite it as

$$ \cos(A/2) = \frac{b+c}{a} \sin(A/2) $$

Multiply both sides by $2 \cos(A/2)$

$$ 2 \cos^2(A/2) = \frac{b+c}{a} \sin(A) $$

Now use the rule $a/\sin(A)=b/\sin(B)=c/\sin(C)$ to transform this to:

$$ 2 \cos^2(A/2) = \sin(B)+\sin(C) $$

Using Simpson's rule on the second hand of the equation to sum the sines gives:

$$ 2 \cos^2(A/2) = 2\sin((B+C)/2)\cos((B-C)/2) $$

Since in a triangle $A+B+C=\pi$ we have $ 2\sin((B+C)/2) = 2\sin((\pi-A)/2) = 2\cos(A/2) $ and thus

$$ \cos(A/2) = \cos((B-C)/2) $$

Or $A+C=B$ (If $B>C$). Combining this with $A+B+C=\pi$ this gives $B=\pi/2$.

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I tried a geometry proof for this and thought it was worth sharing it here.

I am unable to draw the figure. I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures.

Consider the triangle $ABC$. Let the angle bisector of $A$ meet $BC$ at $D$.

By angle bisector theorem, we get

$\frac{AB}{AC} = \frac{DC}{DB}$. Hence, we get $\frac{b}{c} = \frac{DC}{DB}$.

Add one to both sides, we get

$\frac{b+c}{c} = \frac{DC+DB}{DB} = \frac{a}{DB}$.

Rearranging we get $\frac{c}{DB} = \frac{b+c}{a} = \cot(\frac{A}{2})$.

A similar argument (or using the fact that $\frac{c}{DB} = \frac{b}{DC}$) yields, $\frac{b}{DC} = \frac{b+c}{a} = \cot(\frac{A}{2})$.

Now, we are almost done. (Note that if $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$, the above is true.) However, we still need to prove that for no other possibility this is satisfied.

To prove the above. Draw a perpendicular to $AB$ at $B$ and similarly draw a perpendicular to $AC$ at $C$. Let these two will intersect $AD$ at $D'$ and $D''$. Now note that $BD' = BD$ since $\frac{BD}{AB} = \cot(\frac{A}{2}) = \frac{BD'}{AB}$ (Since $D'BA$ is a right angled triangle). Similarly $CD'' = CD$.

But this is not possible unless $D' = D$ or $D'' = D$, which implies $\angle{B} = \frac{\pi}{2}$ or $\angle{C} = \frac{\pi}{2}$

Again, I would appreciate if some one could take the effort to draw the picture or let me know what is the easiest way to draw such pictures. The argument I hope however gives you the idea behind the method.

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