prove $a,b,c$ in A.P if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$
$begingroup$
In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
trigonometry triangle
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
$endgroup$
add a comment |
$begingroup$
In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
trigonometry triangle
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
$endgroup$
add a comment |
$begingroup$
In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
trigonometry triangle
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
$endgroup$
In $Delta ABC$, if $tandfrac{A}{2}=dfrac{5}{6}$ and $tandfrac{C}{2}=dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
sin A=frac{2.5}{6}.frac{36}{61}=frac{60}{61}\
sin C=frac{2.2}{5}.frac{25}{29}=frac{20}{29}\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
trigonometry triangle
trigonometry triangle
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
asked Jan 4 at 16:37
ss1729ss1729
2,01511124
2,01511124
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$
Use
$$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
i was thinking of doing that straightaway .. but is there a better way other than this ?
$endgroup$
– ss1729
Jan 4 at 16:50
add a comment |
$begingroup$
It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
Can you finish now?
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
Hint:
Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,
$$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
You can first deduce
$$
tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
=frac{20}{37}
$$
Therefore
$$
sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
$$
Similarly,
$$
sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
$$
By the sine law,
$$
frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
$$
answered Jan 4 at 17:56
egregegreg
184k1486205
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$
Use
$$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
i was thinking of doing that straightaway .. but is there a better way other than this ?
$endgroup$
– ss1729
Jan 4 at 16:50
add a comment |
$begingroup$
$$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$
Use
$$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$begingroup$
i was thinking of doing that straightaway .. but is there a better way other than this ?
$endgroup$
– ss1729
Jan 4 at 16:50
add a comment |
$begingroup$
$$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$
Use
$$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
$$dfrac1{tandfrac B2}=cot B/2=tan(A/2+C/2)=?$$
Use
$$sin2x=dfrac{2tan x}{1+tan^2x}=?$$ for $2x=A,B,C$
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
edited Jan 4 at 16:50
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 16:48
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
$begingroup$
i was thinking of doing that straightaway .. but is there a better way other than this ?
$endgroup$
– ss1729
Jan 4 at 16:50
add a comment |
$begingroup$
i was thinking of doing that straightaway .. but is there a better way other than this ?
$endgroup$
– ss1729
Jan 4 at 16:50
$begingroup$
i was thinking of doing that straightaway .. but is there a better way other than this ?
$endgroup$
– ss1729
Jan 4 at 16:50
$begingroup$
i was thinking of doing that straightaway .. but is there a better way other than this ?
$endgroup$
– ss1729
Jan 4 at 16:50
add a comment |
$begingroup$
It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
Can you finish now?
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
Can you finish now?
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
add a comment |
$begingroup$
It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
Can you finish now?
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
$endgroup$
It is $$tan(alpha/2)=frac{r}{s-a}$$ and $$tan(gamma/2)=frac{r}{s-c}$$ where $$s=frac{a+b+c}{2}$$ so we get $$frac{5}{6}(s-a)=frac{2}{5}(s-c)$$
Can you finish now?
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
answered Jan 4 at 17:08
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.8k42866
77.8k42866
add a comment |
add a comment |
$begingroup$
Hint:
Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,
$$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,
$$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,
$$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
Hint:
Like In $Delta ABC$, find $cotdfrac{B}{2}.cotdfrac{C}{2}$ if $b+c=3a$,
$$2b=a+c$$ will hold true if $tandfrac A2tandfrac B2=dfrac13$
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 4 at 17:25
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
add a comment |
add a comment |
$begingroup$
You can first deduce
$$
tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
=frac{20}{37}
$$
Therefore
$$
sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
$$
Similarly,
$$
sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
$$
By the sine law,
$$
frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
$$
answered Jan 4 at 17:56
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You can first deduce
$$
tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
=frac{20}{37}
$$
Therefore
$$
sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
$$
Similarly,
$$
sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
$$
By the sine law,
$$
frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
$$
answered Jan 4 at 17:56
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You can first deduce
$$
tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
=frac{20}{37}
$$
Therefore
$$
sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
$$
Similarly,
$$
sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
$$
By the sine law,
$$
frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
$$
answered Jan 4 at 17:56
egregegreg
184k1486205
$endgroup$
You can first deduce
$$
tanfrac{B}{2}=tanleft(frac{pi}{2}-frac{A+C}{2}right)=cotfrac{A+C}{2}=
frac{1-tanfrac{A}{2}tanfrac{C}{2}}{tanfrac{A}{2}+tanfrac{C}{2}}=frac{2/3}{37/30}
=frac{20}{37}
$$
Therefore
$$
sin A=frac{2(5/6)}{1+25/36}=frac{60}{61}
$$
Similarly,
$$
sin B=frac{1480}{1769}qquad sin C=frac{20}{29}
$$
By the sine law,
$$
frac{a+c}{2}=frac{b}{2sin B}(sin A+sin C)=
bfrac{1769}{2960}left(frac{60}{61}+frac{20}{29}right)=b
$$
answered Jan 4 at 17:56
egregegreg
184k1486205
answered Jan 4 at 17:56
egregegreg
184k1486205
answered Jan 4 at 17:56
egregegreg
184k1486205
answered Jan 4 at 17:56
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
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