• Aug 3rd 2006, 07:30 AM
classicstrings
When I work out x I get 2 different answers. Wondering if someone can explain?
• Aug 3rd 2006, 07:42 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
When I work out x I get 2 different answers. Wondering if someone can explain?

http://www.mathhelpforum.com/math-he...ntid=857&stc=1

http://www.mathhelpforum.com/math-he...ntid=858&stc=1

These links don't work so noboby will know what the question/s is/are.

RonL
• Aug 3rd 2006, 07:45 AM
classicstrings
Quote:

Originally Posted by CaptainBlack
These links don't work so noboby will know what the question/s is/are.

RonL

Done. Still dont know why two values of x come up. (it isnt possible!)
• Aug 3rd 2006, 08:31 AM
Quick
Quote:

Originally Posted by classicstrings
Done. Still dont know why two values of x come up. (it isnt possible!)

I don't think sine works unless they're right triangles
• Aug 3rd 2006, 08:33 AM
CaptainBlack
Quote:

Originally Posted by classicstrings
When I work out x I get 2 different answers. Wondering if someone can explain?

Briefly; there is no such triangle if the angle marked as 60 degrees is that,
and the parallel lines are, and the lengths are as indicated, then your angle
marked as 50 degrees will in fact be more like 40 degrees, ie the data you
have been given is inconsistent, so of course you get different answers
depending on how you calculate things.

RonL
• Aug 3rd 2006, 09:37 AM
topsquark
Quote:

Originally Posted by Quick
I don't think sine works unless they're right triangles

For the record, both the Law of Sines and Law of Cosines work on any triangle.

-Dan
• Aug 3rd 2006, 11:38 AM
Soroban
Hello, classicstrings!

The diagram is quite impossible . . . They lied to us!
Code:

              C               *             *  *             *    *           *        *       b=4 *          * a         *              *         *                *       * 60°                *       * * * * * * * * * * * * *       A        c=6          B

Law of Cosines: .$\displaystyle a^2\:=\:b^2 + c^2 - 2bc\cos A$

We have: .$\displaystyle a^2\:=\:4^2 + 6^2 - 2\cdot4\cdot6\cos60^o\:=$ $\displaystyle \:28\quad\Rightarrow\quad a \,= \,\sqrt{28} \,= \,2\sqrt{7}$

Law of Sines: .$\displaystyle \frac{\sin B}{b}\,=\,\frac{\sin A}{a}$

We have: .$\displaystyle \frac{\sin B}{4}\,=\,\frac{\sin60^o}{2\sqrt{7}}\quad \Rightarrow\quad \sin B \:=\:\frac{4\sin60^o}{2\sqrt{7}}\:=\:0.654653671$

Therefore: .$\displaystyle B\;=\;\sin^{-1}(0.654653671)\:=\:$ $\displaystyle 40.839339654 \quad\Rightarrow\quad B \:\approx \:41^o$

See? . . . and they told us that $\displaystyle B\,=\,50^o$