# trigo help

• December 28th 2010, 09:39 AM
prasum
trigo help
prove that if in triangle ABC tan(A/2)=5/6 tan(B/2)=2/5 then a,b,c will be in AP
• December 28th 2010, 10:04 AM
what's "AP"
• December 28th 2010, 10:04 AM
Plato
Quote:

Originally Posted by prasum
prove that if in triangle ABC tan(A/2)=5/6 tan(B/2)=2/5 then a,b,c will be in AP

I see that you have over 100 other postings. So you should understand that this is not a homework service nor is it a tutorial service.
PLease either post some of your own work on this problem or explain what you do not understand about the question.
• December 29th 2010, 04:44 AM
prasum
i have found tan A and Tan B through tan2A formula after that i have found out Tan C by Tan A+TanB+TanC=TanATanBTanC brcause A+b+c=180 degrees after that how should i find a,b,c and prove they are in arithmetic progression
• December 29th 2010, 09:13 AM
wouldn't $sinA=\frac{2tan(A/2)}{1+tan^2(A/2)}$ and $\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}$ work?
• December 30th 2010, 07:18 AM
Quote:

Originally Posted by prasum
prove that if in triangle ABC, tan(A/2)=5/6, tan(B/2)=2/5, then a,b,c will be in AP

$\displaystyle\ tan\left[\frac{A}{2}\right]=\frac{5}{6}\Rightarrow\ sinA=\frac{2tan\left[\frac{A}{2}\right]}{1+tan^2\left[\frac{A}{2}\right]}=\frac{\left[\frac{5}{3}\right]}{\left[\frac{61}{36}\right]}=\frac{60}{61}$

$\displaystyle\ tan\left[\frac{B}{2}\right]=\frac{2}{5}\Rightarrow\ sinB=\frac{\left[\frac{4}{5}\right]}{\left[\frac{29}{25}\right]}=\frac{20}{29}$

$sinC=sin[180^o-(A+B)]=sin(A+B)=sinAcosB+cosAsinB$

We obtain the cosines from Pythagoras' Theorem.

$x^2+60^2=61^2\Rightarrow\ x^2=3721-3600=121\Rightarrow\ x=11$

$\displaystyle\Rightarrow\ cosA=\frac{11}{61}$

$y^2+20^2=29^2\Rightarrow\ y^2=29^2-20^2=441\Rightarrow\ y=21$

$\displaystyle\Rightarrow\ cosB=\frac{21}{29}$

$\displaystyle\ sinC=\frac{60}{61}\;\frac{21}{29}+\frac{11}{61}\;\ frac{20}{29}=\frac{1260+220}{1769}=\frac{1480}{176 9}$

From the Law of Sines

$\displaystyle\frac{a}{sinA}=\frac{b}{sinB}=\frac{c }{sinC}$

$\Rightarrow\ a:b:c=sinA:sinB:sinC$

If a, b, and c are in Arithmetic progression,
then the sum of the 3 sides is 3 times the "middle side".

$p+(p+d)+(p+2d)=3p+3d=3(p+d)$

Hence, the sum of the 3 sines must be 3 times the sum of the "middle sine".

$sinA+sinB+sinC=2.50989259469$

$3sinC=2.50989259469$

Therefore

a, c, b are in Arithmetic Progression.
b, c, a are in Arithmetic Progression.