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Using $$\tan \frac A2 +\tan \frac B2 +\tan \frac C2=1-\tan \frac A2 \tan \frac B2 \tan \frac C2$$ Substituting the given values $$\frac 56 +\frac 25 +\tan \frac B2=1-\frac 56 .\frac 25 \tan \frac B2$$

$$\tan \frac B2=\frac{-7}{40}$$ But I don’t know what to do with this information. How do I prove they are in GP?

N. F. Taussig
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Aditya
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  • What are $a,b,c$ ? – The Demonix _ Hermit Dec 02 '19 at 13:32
  • Be careful: are $;a,b,c;$ the same as $;A,B,C;$ ? Use the same symbols for the same things all the time. And what is $;c;$ ...or $;C;$, anyway? – DonAntonio Dec 02 '19 at 13:36
  • You mean ABC is a triangle – mathsdiscussion.com Dec 02 '19 at 13:39
  • @DonAntonio yes I am sorry I forgot to mention. The sides and angles are according to standard notation ie. a is opposite A – Aditya Dec 02 '19 at 13:49
  • Where did you get the first equation? It seems to me it is not true for a general triangle with interior angles $A, B,C.$ – David K Dec 02 '19 at 13:56
  • If $C$ is an interior angle then we should find that $0 < C/2 <\pi/2$ and $\tan(C/2)>0.$ So the fact you get a negative number is an indication something is wrong. – David K Dec 02 '19 at 14:04
  • @DavidK since $A+B+C=\pi$, then $\frac{A+B+C}{2}=\frac {\pi}{2}$. Then $\tan \pi/2 =\frac 10$ and then I used the formula for $\tan {P+Q+R}$ – Aditya Dec 02 '19 at 15:06
  • See https://math.stackexchange.com/questions/3450441/in-a-triangle-if-tana-2-tanb-2-tanc-2-are-in-arithmetic-progre/3451324?r=SearchResults#3451324 – lab bhattacharjee Dec 02 '19 at 15:09
  • @DavidK I realised that the number shouldn’t be negative, but unless there is something wrong with my arithmetic, I don’t see a problem – Aditya Dec 02 '19 at 15:10
  • @Aditya - It appears that the sides are in arithmetic progression, not geometric. – Quanto Dec 02 '19 at 15:22
  • First, you need to be careful anytime you write $1/0.$ It's not clear if you did the sum-of-three-angles formula correctly (see https://math.stackexchange.com/questions/177640/prove-tanaby-frac-tan-a-tan-b-tan-y-tan-a-tan-b-tan-y1-tan-a-tan), but if you got your formula by setting the numerator equal to $1,$ that's a mistake. Your final formula is clearly incorrect for an equilateral triangle, so we know there's a mistake somewhere. – David K Dec 02 '19 at 17:54

5 Answers5

1

Your question appears to be wrong. Or my arithmetic could be, but I keep staring at it looking for the reason why.

Since $\sin A=\frac{2\times\frac56}{1+\frac{25}{36}}=\frac{60}{61}$ and similarly $\sin B=\frac{20}{29},\,\cos A=\frac{11}{61},\,\cos B=\frac{21}{29}$,$$\sin C=\sin(A+B)=\sin A\cos B+\cos A\sin B=\frac{1480}{61\times 29}.$$By the sine rule, what matters is whether the sines of $A,\,B,\,C$ are in geometric progression. Well, they're not.

J.G.
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From the given $\tan \frac A2 = \frac56$ and $\tan \frac B2 = \frac25$, we have

$$\sin A = \frac{2\tan \frac A2}{1+\tan^2 \frac A2}=\frac{60}{61}, \>\>\>\>\>\cos A = \frac{11}{61}$$

$$\sin B = \frac{2\tan \frac B2}{1+\tan^2 \frac B2}=\frac{20}{29}, \>\>\>\>\>\cos B = \frac{21}{29}$$

Then,

$$\frac{a+b}{2c}=\frac{\sin A +\sin B}{2\sin (A+B)} =\frac{\sin A +\sin B}{2(\sin A\cos B + \cos A\sin B)}$$ $$=\frac{\frac{60}{61} +\frac{20}{29}} {2(\frac{60}{61}\cdot\frac{21}{29}+ \frac{11}{61} \cdot\frac{20}{29})}=\frac{60\cdot29+61\cdot20 }{2(60\cdot21+11\cdot20)}=1$$

Thus, $a$, $b$ and $c$ are in arithmetic progression (not geometric).

Quanto
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Use $$\tan(\alpha/2)=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$$ $$\tan(\beta/2)=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$$ since we have $$\tan(\pi/2-\frac{\gamma}{2})=\frac{\frac {5}{6}+\frac{2}{5}}{1-\frac{5}{6}\times\frac{2}{5}}$$ we get $$\cot(\gamma/2)=\frac{20}{37}$$ so $$\tan(\gamma/2)=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$ Solving this system for $a,b,c$ we get $b=\frac{61}{87}a,c=\frac{74}{87}a$

Dr. Sonnhard Graubner
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$Cos A=\frac{1- tan^2(A/2)}{1+tan^2(A/2)}=\frac{11}{61}$

$Cos B=\frac{1- tan^2(B/2)}{1+tan^2(B/2)}=\frac{21}{29}$

$$ a^2=b^2+c^2-2bc Cos A$$

$$ b^2=a^2+c^2-2ac Cos B$$

Summing these relations we get:

$$2 c^2=2ac Cos B +2bc Cos A$$

Plugging values of (Cos A) and (cos B) we get:

$$1189 c=319 b+ 1281 a$$

It does not show any geometric progression.

sirous
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I would like to derive how the values of $\tan\dfrac A2,\tan\dfrac B2$ are decided

Method$\#1:$

Like Quanto's method $$\dfrac{a+b}{2c}=\dfrac{\sin A+\sin B}{2\sin C}=\dfrac{2\sin\dfrac{A+B}2\cos\dfrac{A-B}2}{4\sin\dfrac C2\cos\dfrac C2}=\dfrac{\cos\dfrac{A-B}2}{2\cos\dfrac{A+B}2}=\dfrac{\cos\dfrac A2\cos\dfrac B2+\sin\dfrac A2\sin\dfrac B2}{2\left(\cos\dfrac A2\cos\dfrac B2-\sin\dfrac A2\sin\dfrac B2\right)}$$

If this $=1,$ $$3\sin\dfrac A2\sin\dfrac B2=\cos\dfrac A2\cos\dfrac B2\iff\tan\dfrac A2\tan\dfrac B2=\dfrac13$$

Method$\#2:$

Here In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$,

we have derived $\sin A,\sin B,\sin C$ will be in Arithmetic Progression iff $\cot\dfrac A2,\cot\dfrac B2, \cot\dfrac C2$ are so.

Using In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2),

$$\cot\dfrac A2+\cot\dfrac B2+\cot\dfrac C2=\cot\dfrac A2\cot\dfrac B2\cot\dfrac C2$$

Now if $\cot\dfrac A2+\cot\dfrac B2=2\cot\dfrac C2,$

$$3\cot\dfrac C2=\cot\dfrac A2\cot\dfrac B2\cot\dfrac C2$$

As $\cot\dfrac C2\ne0,$ the result follows.

lab bhattacharjee
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