# Right Triangles

• August 25th 2008, 05:25 PM
topaz192
Right Triangles
Am I doing the right steps for this?

The question is: solve each triangle as described.

c= 9.5 b= 3.7

http://i20.photobucket.com/albums/b2...e_188/math.jpg

Do i do the pygahthom therom [sry spelln] first or can i just go ahead and fing the sides by doing 90 - one of the sides?

I just did 9.5^2 - 3.7^2 = 669.89

What do I do now??(Doh)
• August 25th 2008, 05:29 PM
skeeter
$a = \sqrt{c^2 - b^2}$
• August 25th 2008, 05:35 PM
Jhevon
Quote:

Originally Posted by topaz192
Am I doing the right steps for this?

The question is: solve each triangle as described.

c= 9.5 b= 3.7

http://i20.photobucket.com/albums/b2...e_188/math.jpg

Do i do the pygahthom therom [sry spelln]

since you spelled "sorry" as "sry" and "spelling" as "spelln", i guess spelling wasn't a priority for you in the first place

Quote:

I just did 9.5^2 - 3.7^2 = 669.89

What do I do now??(Doh)
you are on the right track, but i don't know what happened. how exactly did you get 669.89? just looking at the expression we know that it has to be less than 100. now note that the answer you get is $a^2$. so how do we find $a$?

then recall the required trig ratios. pick your fancy, you only need to find one angle, then the other is easy and no trig ratios are required

$\text{sine } = \frac {\text{opposite}}{\text{hypotenuse}}$

$\text{cosine } = \frac {\text{adjacent}}{\text{hypotenuse}}$

$\text{tangent } = \frac {\text{opposite}}{\text{adjacent}}$
• August 25th 2008, 05:38 PM
topaz192
Ok i got 76.56 and the square root would be: 8.75

So do i just do it as

TAN a = 3.7 / 9.5 and get 21.28?

Is that all?
• August 25th 2008, 05:40 PM
topaz192
Quote:

Originally Posted by Jhevon
since you spelled "sorry" as "sry" and "spelling" as "spelln", i guess spelling wasn't a priority for you in the first place

lol i realized that afterwards, sorry it's a habit. (Nod)