Finding the height of this triangle

In summary, 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) (
  • #1
sh86
19
0
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?
 
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  • #2
Your mistake is here: tan40 = h/x (so x = (tan40)/h)

x should be equal to h / tan40
 
  • #3
You correctly say

[tex] \tan 40 = \frac h x [/tex]

try solving this for x again. Perhaps you will find a different result. :smile:
 
  • #4
LOL! I combed over my solution about a thousand times looking for what I did wrong and I couldn't see that. Thanks guys.
 
  • #5
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

[tex]\Theta = 180 - (40+47)[/tex]

[tex]\Theta = 93\cdot[/tex]

then apply sin law

[tex]\frac{125}{sin93} = \frac{x}{sin40}[/tex]

[tex]125*sin40 = x*sin93[/tex]

[tex]\frac{125*sin40}{sin93} = x[/tex]

[tex]80.5 = x[/tex]

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
[tex]sin47 = \frac {h}{80.5}[/tex]
 
  • #6
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.
 

1. How do you find the height of a triangle?

To find the height of a triangle, you can use the formula h = (2 x area)/base. This means that the height is equal to twice the area divided by the base of the triangle.

2. Can you find the height of a triangle without knowing its base?

No, the height of a triangle cannot be determined without knowing the base. The base and height of a triangle are directly related and both are needed to calculate the area of the triangle.

3. What tools do you need to find the height of a triangle?

To find the height of a triangle, you will need a measuring tool, such as a ruler or protractor, to measure the base and sides of the triangle. You will also need a calculator to perform the necessary calculations.

4. Can you use trigonometry to find the height of a triangle?

Yes, you can use trigonometry to find the height of a triangle. You can use the trigonometric ratios of sine, cosine, and tangent to find the height, depending on the given information about the triangle.

5. Are there any special cases when finding the height of a triangle?

Yes, there are some special cases when finding the height of a triangle. For example, in a right triangle, the height is equal to one of the sides, and in an equilateral triangle, the height is equal to the square root of three divided by two times the length of one of the sides.

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