Finding the height of this triangle

[sh86]

The problem is to find h on this triangle:

http://img215.imageshack.us/img215/7006/triangleum3.png

With the help of the law of sines I've already finished this problem. But I tried doing it a different way and my new solution isn't working and I can't figure out why. Here's what I did:

(1) Break the bottom part like so

http://img221.imageshack.us/img221/3079/bottomba3.png

(2) Use tan to solve for x in two ways

tan40 = h/x (so x = (tan40)/h)
tan47 = h/(125-x) (so x = 125 - hcot47)

(3) Since both those equations equal x, they equal each other, so I do this:

(tan40)/h = 125 - hcot47

Multiplying both sides by h, blah blah blah, I get this

h²cot47 - 125h + tan40 = 0.

So I have a quadratic equation in h. When I use the quadratic formula and do all the solving (which I'll omit since it would take a lot of space to write) I get an incorrect answer. Where did I go wrong?

 

[chanvincent]

Your mistake is here: tan40 = h/x (so x = (tan40)/h)

x should be equal to h / tan40

 

[Integral]

You correctly say

\tan 40 = \frac h x

try solving this for x again. Perhaps you will find a different result.

 

[sh86]

LOL! I combed over my solution about a thousand times looking for what I did wrong and I couldn't see that. Thanks guys.

 

[snowJT]

The way I would of done it would of been like so...

I wouldn't of broken up 125 to x and 125-x.. I would of found one of the other sides using sin law because we can figure out the missing angle

\Theta = 180 - (40+47)

\Theta = 93\cdot

then apply sin law

\frac{125}{sin93} = \frac{x}{sin40}

125*sin40 = x*sin93

\frac{125*sin40}{sin93} = x

80.5 = x

then you have the right triangle which is a right angel triangle with the hypotinuse and a given angel.. that's all you need to find h using
sin47 = \frac {h}{80.5}

 

[Gib Z]

No, he said he already did it that way, But did it another way and had a problem with that. Nice bit of tex though.