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Right Triangles within a right triangle. (Trig)

  1. Jun 21, 2008 #1
    "To Estimate the height of a mountain above a level plain, the angle of elevation to the top of the mountain is measured to be 32 degrees. One thousand feet closer to the mountain along the plain, it is found that the angle of eleveation is 35 degrees. Estimate the height of the mountain."

    First I drew the picture out and constructed a straight vertical line 1000ft from the vertex. I labeled the unknown part of the triangle's base as x-1000. From there I can find all the side lengths of the small right triangle using the trigonometric ratios. I know that the smallest right triangle is similar to the the one with side x, but I don't know how to apply this in finding the height (y), of the right triangle (mountain).
  2. jcsd
  3. Jun 21, 2008 #2
    1. Find all the angles for the two inner triangles.
    2. Then use the sine rule. There's several ways to solve for the height.
    3. Also, could you post a picture because I do not think you're drawing the diagram correctly. The two triangles are not similar because the angles are different i.e. 32 degrees and 35 degrees.

    There are three triangles.

    1. The biggest triangle with the height, and the longest hypotenuse and the angle 32 degrees.
    2. The left most inner triangle (which is not a right triangle) with a base side of 1,000 ft and the angle 32 degrees.
    3. The right most inner triangle (which is a right triangle) with angle 35 degrees and a base of x-1000 ft, where x is the base side of the largest triangle.
    Edit: Yes, you are not drawing it correctly, since it says that "One thousand feet closer to the mountain along the plain" the 1000 ft should be along the plain i.e. horizontally not a "straight vertical line 1000ft from the vertex".
    Last edited: Jun 21, 2008
  4. Jun 21, 2008 #3
    Yeah, I think its confusing the way i said it. But I don't have the technology with me for a picture. I would think that there would be one more triangle...implied.

    if the vertex was the 32 degree angle and I moved horizontally 1000 ft, and plotted that with a point on the base. And drew a line vertically from that angle until it intersects the hypotenuse of the larger triangle, it should create a triangle with 32 degree angle and base of 1000 ft. Shouldn't this triangle be similar with the bigger triangle with angle 32 degrees and base x?

    Give me a moment, I'm going to try the problem again...
  5. Jun 21, 2008 #4
    Here's a diagram that I made.
  6. Jun 22, 2008 #5


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    The picture is not the point- do what konthelion suggested:

    What are the angles of that smaller inside triangle? You are already given one and it's easy to find the other two. With that information, you can use the "sine law" to find the two unknown lengths. Now, you know the length of the hypotenuse of the inner right triangle and so can use trig functions to find the height.
  7. Jun 22, 2008 #6
    ok, I believe I got it.

    I used the law of sines (which i just learned) and found the hypotenuse of the 35 degree right angle by using a/sinA=b/sinB=c/sinC. Since c/SinC is 1000/sin3, I used this as a ratio to find the length of side a;

    a/sin32 = 1000/sin3,

    a=(1000sin32)/sin3 = 10,125.33834

    then I find the length of y by sin35 = y/10,125 and found y to equal approximately 5,807.655ft.

    Look correct?
  8. Jun 22, 2008 #7
    I just checked the answer in the back of the book, and I was right on.

    Thanks, especially to konthelion for his time.
  9. Jun 22, 2008 #8
    Cool. That's the same answer I ended up with, too.
  10. Jan 12, 2009 #9
    This will also work

    h = (1000 * tan32 * tan35) / ( tan35 - tan32)
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