It is only multiples of 3, I was wondering if someone could please look over this? I am having an extremely hard time with these...i did not get #33...please help

http://i103.photobucket.com/albums/m...91919/22-1.jpg

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- Oct 30th 2007, 01:00 PMaikenfanSpecial Right Triangles
It is only multiples of 3, I was wondering if someone could please look over this? I am having an extremely hard time with these...i did not get #33...please help

http://i103.photobucket.com/albums/m...91919/22-1.jpg - Oct 30th 2007, 01:07 PMTKHunny
Are you CERTAIN you have all the labels correct? It is rather confusing to use lower case on the drawing and upper case in the problem set.

1) ALWAYS remember that drawings may not be to scale. Do NOT let your eyes decieve you.

2) 30-60-90 ALWAYS has the long leg equal to $\displaystyle \sqrt{3}$ times the short leg.

Short = $\displaystyle 8$, Long = $\displaystyle 8\sqrt{3}$

Long = $\displaystyle 25$, Short = $\displaystyle \frac{25}{\sqrt{3}}$ - Oct 30th 2007, 01:55 PMaikenfan
Yes, the diagrams came labeled...did I do something wrong?

- Oct 30th 2007, 01:59 PMSoroban
Hello, aikenfan!

They're fine . . . except for #33 . . .Code:`*`

* *

c * 60° *

* * a

* 30° *

* * * * * * * *

b

33) & \_\_ & 25 & \_\_ \end{array} $

For this triangle $\displaystyle a:b:c$ is in the ratio: .$\displaystyle x:x\sqrt{3}:2x$

We are told: .$\displaystyle b = 25$

. . So we have: .$\displaystyle x\sqrt{3} \,=\,25\quad\Rightarrow\quad x \:=\:\frac{25}{\sqrt{3}} \:=\:\frac{25\sqrt{3}}{3} $

Therefore: .$\displaystyle \begin{array}{ccccc}a & = & x & = & \frac{25\sqrt{3}}{3} \\ c & = & 2x & = & \frac{50\sqrt{3}}{3}\end{array}$

- Oct 30th 2007, 02:07 PMaikenfan
Thank you very much for all of your help

:)