Special Right Triangles

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October 30th 2007, 01:00 PM
aikenfan

Special 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
October 30th 2007, 01:07 PM
TKHunny

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 $\sqrt{3}$ times the short leg.

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

Long = $25$ , Short = $\frac{25}{\sqrt{3}}$
October 30th 2007, 01:55 PM
aikenfan

Yes, the diagrams came labeled...did I do something wrong?
October 30th 2007, 01:59 PM
Soroban

Hello, aikenfan!

They're fine . . . except for #33 . . .

Code:

* * * c * 60° * * * a * 30° * * * * * * * * * b

$\begin{array}{cccc} & a & b & c \\<br /> 33) & \_\_ & 25 & \_\_ \end{array}$

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

We are told: . $b = 25$
. . So we have: . $x\sqrt{3} \,=\,25\quad\Rightarrow\quad x \:=\:\frac{25}{\sqrt{3}} \:=\:\frac{25\sqrt{3}}{3}$

Therefore: . $\begin{array}{ccccc}a & = & x & = & \frac{25\sqrt{3}}{3} \\ c & = & 2x & = & \frac{50\sqrt{3}}{3}\end{array}$
October 30th 2007, 02:07 PM
aikenfan

Thank you very much for all of your help
:)