# Seemingly flawed algebra

• Mar 19th 2013, 06:18 PM
Paze
Seemingly flawed algebra
Riddle me this MHF.

How come the algebra doesn't provide me with two solutions?

(SinX = SinB and 49 is 49°. This was sketched in a hurry..More like doodled..)

http://s18.postimage.org/xe1gohgqx/algebraproblem.png
• Mar 19th 2013, 06:38 PM
Shakarri
Re: Seemingly flawed algebra
Quote:

Originally Posted by Paze
Riddle me this MHF.

This was sketched in a hurry..More like doodled..)

I'd suggest using a ruler next time ;)

• Mar 19th 2013, 06:43 PM
Paze
Re: Seemingly flawed algebra
Quote:

Originally Posted by Shakarri
I'd suggest using a ruler next time ;)

Yes, but the general idea should be obvious. I have two triangles with 2 identical sides and 1 identical corner. That means that I have a different corner B in each triangle but solving with algebra only yields one solution. I am supposed to somehow 'know' to subtract the 2nd solution from 180 to get my 2nd corner and we all know that that's not how mathematics work. There is a reason and I would very much like to know it.

• Mar 19th 2013, 07:23 PM
Shakarri
Re: Seemingly flawed algebra
Quote:

Originally Posted by Paze

Normally when people say "riddle me this" they mean the question as a joke so I gave a non serious answer.
sin(B)=sin(B+90)=sin(B+450)
• Mar 19th 2013, 07:26 PM
Paze
Re: Seemingly flawed algebra
Quote:

Originally Posted by Shakarri
Normally when people say "riddle me this" they mean the question as a joke so I gave a non serious answer. I think that the angle in the second triangle is B+90 since sin(B)=sin(B+90)=sin(B+450).

Oh you answered my question indirectly.

It's because sin(180-B)=sin(b).
(Not b+90)

Thanks!
• Mar 19th 2013, 07:31 PM
Shakarri
Re: Seemingly flawed algebra
Oops that was silly of me. I tried sketching it and got 105 and 75 :)