problem triangle

jamaica25

New member
Joined
Mar 5, 2015
Messages
6
this is a problem I saw on this forum but cannot find the answer so I am posting a new thread

cos a= tan b find sin a ??
cos b = tan c
cos c =tan a

using substitution sin a =cos. ^2a*sin c/(sin b)

I would think you would want degree answer

sin a = 22.5 degree by trial and error

is there such a triangle ??
 
this is a problem I saw on this forum but cannot find the answer so I am posting a new thread

cos a= tan b find sin a ??
cos b = tan c
cos c =tan a

using substitution sin a =cos. ^2a*sin c/(sin b)

I would think you would want degree answer

sin a = 22.5 degree by trial and error

is there such a triangle ??
As I implied on the other thread
http://www.freemathhelp.com/forum/t...nyone-solve-this-equation?highlight=cosa=tanb
you might replace those quantities involving, say c, by using
c = \(\displaystyle \pi\)-(a+b)
and see if that helps.
 
to ishuda


I tried that got nowhere

I don't know what answer to expect

angle a= 22.5 deg.
b =42.83
c =114.76
seems to work
 
to ishuda


I tried that got nowhere

I don't know what answer to expect

angle a= 22.5 deg.
b =42.83
c =114.76
seems to work
Just a trial post with nothing else said.
to ishuda


I tried that got nowhere

I don't know what answer to expect

angle a= 22.5 deg.
b =42.83
c =114.76
seems to work
Not quite. cos(c) is negative, tan(a) is positive. I believe there is no solution. Since these equations are circular, i.e. replace a with b, b with c, and c with a and you have the same equations, the following applies for all three variables:
\(\displaystyle cos^2(a) + 1 = tan^2(b) + 1 = \dfrac{1}{cos^2(b)} = \dfrac{1}{tan^2(c) + 1 - 1}\)
\(\displaystyle =\dfrac{1}{\frac{1}{cos^2(c)} - 1} = \dfrac{cos^2(c)}{1 - cos^2(c)}=\dfrac{tan^2(a)}{1-tan^2(a)}\)
\(\displaystyle =\dfrac{sin^2(a)}{cos^2(a)-sin^2(a)}=\dfrac{1 - cos^2(a)}{2 cos^2(a) - 1}\)
or, simplifying,
\(\displaystyle 2 cos^4(a) + 2 cos^2(a) - cos^2(a) - 1 = 1 - cos^2(a)\)
or
\(\displaystyle cos^4(a) + cos^2(a) - 1 = 0\)
which gives
\(\displaystyle cos^2(a) = \dfrac{-1 \pm \sqrt{5}}{2}\)
since the minus sign gives complex roots which, for this problem, are not allowed we have
cos(a)=\(\displaystyle \pm \sqrt{0.5 (\sqrt{5} - 1)}\)
and a (and b and c) must be given by
a ~ 2.475353 [~141.827 deg] or a ~ 0.666239 [~38.173 deg]
plus or minus, of course, some number of complete cycles, i.e \(\displaystyle \pm 2 n \pi\). I don't think you can make a combination of those angles, all positive and less than 180 degrees individually, to add to \(\displaystyle \pi\)(180 degrees).
 
SPIT! The only way I seemed to be able to post is to post a reply is to start a new reply with no quote and it keeps changing. I can't edit the other post. Anyone else having problems?

If \(\displaystyle \alpha = Cos^{-1}(\sqrt{0.5 ( -1 + \sqrt{5}}\,\,\,\,);\,\, 0\, \le\, \alpha\, \le\, \frac{\pi}{2}\), then
a=\(\displaystyle \alpha;\, \pi\, -\, \alpha;\, \pi\, +\, \alpha;\, or\, 2\, \pi\, -\, \alpha\, \pm\, 2\, n\, \pi,\, n=0, 1, 2, ...\)
 
SPIT! The only way I seemed to be able to post is to post a reply is to start a new reply with no quote and it keeps changing. I can't edit the other post. Anyone else having problems?
I've had this happen occasionally, too. I've never been able to figure out any rhyme or reason to it. Sometimes, I end up doing my one reply in two or even three separate messages. Sorry I can't help. :oops:
 
to Denis

on this site sim 1512
What is "this site sim 1512"?

can anyone solve this equation 3/01/15
But "3/01/15" is not an "equation"; it's a date. What do you mean by this? :confused:

to ishuda

I tried that got nowhere
Please reply SHOWING what you have done. Thank you.

I don't know what answer to expect
Since, according to you, this cryptic question asked for the value of the sine of angle a, then probably the answer will be a numerical value between -1 and +1.

Are you not familiar with trigonometry? (Are you ever going to provide us with a link to whatever was the FULL text of the actual original exercise?) ;)
 
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