Triangle XYZ has YZ = 3 cm, ZX = 7cm, and m(angle Y) = 60°

G

Guest

Guest
Triangle XYZ has YZ = 3 cm, ZX = 7cm, and m(angle Y) = 60°

XYZ is a triangle in which YZ = 3 cm, ZX = 7 cm and angle Y is 60°. Prove that exact value of cos C is -1/7
 
uhhh ... where is angle "C" in triangle XYZ :?: :shock:
 
It would help considerably if you included a picture.

Draw it in Paint, save it, use http://www.imageshack.us/ to host it, copy and paste the bottom link into the forum. Put on either side of the URL after you paste.
 
Re: Triangle

Hello, atomos!

A silly typo . . . didn't you Preview your post?
Or look at it after you posted it?

I have a circuitous solution . . . I'm sure there's a more elegant one.

\(\displaystyle XYZ\) is a triangle in which \(\displaystyle YZ\,=\,3\) cm, \(\displaystyle ZX\,=\,7\) cm and \(\displaystyle \angle Y \,=\,60^0\)

Prove that exact value of \(\displaystyle \cos\) Z\(\displaystyle \,=\,-\frac{1}{7}\)
Click to expand...
Code: 
                              Z
                              *
                          *    *
                  7   *         * 3
                  *              *
              *                   *
          *                    60° *
    X * * * * * * * * * * * * * * * * Y

Note: \(\displaystyle X,\,Y,\,Z\) are the angles.

\(\displaystyle Z \:=\:180^o\,-\,60^o \,- \,X \:=\:120^o\,-\,X)\)

Then: \(\displaystyle \,\cos Z \;= \;\cos120^o\cos X \,+\,\sin120^o\sin X\)

. . . . . \(\displaystyle \L\cos Z \;= \;\left(-\frac{1}{2}\right)\cos X\,+\,\left(\frac{\sqrt{3}}{2}\right)\sin X\;\) [1]


Law of Sines: \(\displaystyle \,\frac{\sin X}{3}\:=\:\frac{\sin60^o}{7}\;\;\Rightarrow\;\;\sin X \:=\:\frac{3\cdot\sin60^o}{7} \:=\:\frac{3\sqrt{3}}{14}\)

Then: \(\displaystyle \,\cos X \:=\:\sqrt{1\,-\,\sin^2X} \:=\:\sqrt{1\,-\,\left(\frac{3\sqrt{3}}{14}\right)^2} \:=\:\sqrt{\frac{169}{196}} \:=\:\frac{13}{14}\)


We have: \(\displaystyle \,\cos X\:=\:\frac{13}{14},\;\;\sin X\:=\:\frac{3\sqrt{3}}{14}\)

Substitute into [1]: \(\displaystyle \L\,\cos Z \;= \;\left(-\frac{1}{2}\right)\left(\frac{13}{14}\right)\,+\,\left(\frac{\sqrt{3}}{2}\right)\left(\frac{3\sqrt{3}}{14}\right) \;=\;-\frac{13}{28}\,+\,\frac{9}{28}\;=\;-\frac{4}{28}\)

Therefore: \(\displaystyle \L\,\cos Z \:=\:-\frac{1}{7}\)

 
You must log in or register to reply here.
Top