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Math Help - Triangle 2

  1. #1
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    Triangle 2

    I need the method of these... For Q2, no matter how many times I do, I got 9.57 for VB...

    1. Each edge of the uniform tetrahedron ABCD has the length 4a. The point P lies on BC such that BP=3a. Find the cosine of the angle APD.

    i) show that the perpendicular distance of the corner A from the plane BCD is \frac{4}{3}a\sqrt{6}, and hence
    ii) show that the angle between the line AP and the plane BCD is sin^{-1}\sqrt{\frac{32}{39}}

    2. The horizontal base of a pyramid is an equilateral triangle ABC with each side 10m long. The vertex V of the pyramid is at a height of \frac{10\sqrt{6}}{3}m above the base. Find VB.

    i) If the point P lies on VB such that VP= \frac{1}{5}VB, find AP.

    Answer:
    1. \frac{5}{13}
    2. VB=10m, AP= 2\sqrt{21}m
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  2. #2
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    Quote Originally Posted by cloud5 View Post
    ...

    2. The horizontal base of a pyramid is an equilateral triangle ABC with each side 10m long. The vertex V of the pyramid is at a height of \frac{10\sqrt{6}}{3}m above the base. Find VB.

    i) If the point P lies on VB such that VP= \frac{1}{5}VB, find AP.

    Answer:
    1. \frac{5}{13}
    2. VB=10m, AP= 2\sqrt{21}m
    1. I assume that the vertex of the pyramid is vertically above the center M of the equilateral triangle. (In an equilateral triangle the orthocenter, the centroid, the common point of the three angle bisectors,... is the same point which I call here the center of the triangle)

    2. The point M lies on the median through B. The distance |\overline{BM}| = \frac23 \cdot (length\ of\ median)

    The length of the median is

    |m| = \sqrt{10^2-5^2}=5 \sqrt{3}

    Thus |\overline{BM}| = \frac23 \cdot 5 \sqrt{3}

    3. Now use Pythgorean theorem to calculate

    |\overline{VB}| = \sqrt{ \left(\frac23 \cdot 5 \sqrt{3} \right)^2 + \left(\frac{10\sqrt{6}}{3}  \right)^2 } = \sqrt{\frac{100}3 + \frac{200}3}

    Since VB = 10 the pyramid is a tetrahedron.

    4. I'll leave the rest for you.
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    Quote Originally Posted by earboth View Post

    I'll leave the rest for you.
    Do you know how to answer Q1?
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    Quote Originally Posted by earboth View Post

    2. The point M lies on the median through B. The distance |\overline{BM}| = \frac23 \cdot (length\ of\ median)

    The length of the median is

    |m| = \sqrt{10^2-5^2}=5 \sqrt{3}

    Thus |\overline{BM}| = \frac23 \cdot 5 \sqrt{3}

    3. Now use Pythgorean theorem to calculate

    |\overline{VB}| = \sqrt{ \left(\frac23 \cdot 5 \sqrt{3} \right)^2 + \left(\frac{10\sqrt{6}}{3}  \right)^2 } = \sqrt{\frac{100}3 + \frac{200}3}

    Since VB = 10 the pyramid is a tetrahedron.

    4. I'll leave the rest for you.
    Can you sketch the triangle for me? I'm getting more confused... and about this |\overline{BM}| = \frac23 \cdot (length\ of\ median) are you using the vector formula? why there is \frac{2}{3}? I'm sorry if I'm asking too much...
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  5. #5
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    Quote Originally Posted by cloud5 View Post
    Can you sketch the triangle for me? <<<<<<<< yes
    I'm getting more confused... and about this |\overline{BM}| = \frac23 \cdot (length\ of\ median) are you using the vector formula? <<<<<< no
    why there is \frac{2}{3}? I'm sorry if I'm asking too much...
    1. See attachment.

    2. As I've stated previously the point M is simultaneously the point of intersection of the angle bisectors, the heights and the medians.

    3. The point M divides each median in triangle in the ratio 2:1 from the vertex. Thus the line segment \overline{BM} must be \tfrac23 of the complete median.

    4. The sketch was made for another question so the labeling has nothing to do with your problem.
    Attached Thumbnails Attached Thumbnails Triangle 2-tetraeder.png  
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  6. #6
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    Quote Originally Posted by cloud5 View Post
    I need the method of these... For Q2, no matter how many times I do, I got 9.57 for VB...

    1. Each edge of the uniform tetrahedron ABCD has the length 4a. The point P lies on BC such that BP=3a. Find the cosine of the angle APD.

    ...

    ...
    1. You are dealing with 3 triangles:

    \Delta DPC: Calculate the length of \overline{DP} and \overline{AP} using Cosine rule:

    |\overline{DP}| = \sqrt{a^2+16a^2-2 \cdot a \cdot 4a \cdot \frac12} = a\sqrt{13}

    |\overline{AP}| = \sqrt{16a^2+9a^2-2 \cdot 4a \cdot 3a \cdot \frac12} = a\sqrt{13}

    2. Calculate the Cosine of the angle \angle(DPA):

    \cos(\angle(DPA)) = \dfrac{16a^2-13a^2-13a^2}{-2 \cdot a\sqrt{13} \cdot a\sqrt{13}} = \dfrac{-10a^2}{-26a^2} = \dfrac5{13}
    Attached Thumbnails Attached Thumbnails Triangle 2-tetraeder_innenwinkel.png  
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  7. #7
    Junior Member BugzLooney's Avatar
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    So you are using cosine rule a^2+b^2 -2abCosC, only how did you know C was 1/2? Also where did the 9 come from in the second one?
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  8. #8
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    Quote Originally Posted by BugzLooney View Post
    So you are using cosine rule a^2+b^2 -2abCosC, only how did you know C was 1/2? Also where did the 9 come from in the second one?
    1. The "faces" of a tetrahedron are equilateral triangles. Each interior angle in an equilateral triangle is 60°. And \cos(60^\circ) = \tfrac12.

    2. The 2nd equation calculates the length of \overline{AP} which belongs to the triangle \Delta ABP. The known sides of this triangle are: |\overline{AB}|=4a and |\overline{BP}|=3a. And when I square 3a I'll get 9aČ.
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  9. #9
    Junior Member BugzLooney's Avatar
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    Ah I see didn't see the 3a. I see how it is that for part 1. Which law was used to generate the cosine of DPA? Just the law is all I'm looking for so I can work it myself.
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  10. #10
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    Quote Originally Posted by BugzLooney View Post
    Ah I see didn't see the 3a. I see how it is that for part 1. Which law was used to generate the cosine of DPA? Just the law is all I'm looking for so I can work it myself.
    That's the Cosine rule, solved for \cos(\angle DPA).

    In general:

    c^2 = a^2+b^2-2\cdot a\cdot b \cdot \cos(\gamma)~\implies~\boxed{\cos(\gamma)=\dfrac{c  ^2-a^2-b^2}{-2\cdot a\cdot b}}
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