- #1
sh86
- 19
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The problem is to find h on this triangle:
http://img215.imageshack.us/img215/7006/triangleum3.png [Broken]
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 [Broken]
(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?
http://img215.imageshack.us/img215/7006/triangleum3.png [Broken]
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 [Broken]
(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?
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