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Math Help - finding height and angle?

  1. #1
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    finding height and angle?

    hi, i need to try and find the angle of VAC and find the height of this pyramid!

    finding height and angle?-pyramid.jpg

    to find the height of it would i do:
    1/3x20x15= 100??

    as for finding the angle of VAC i don't know what to do!

    if you can help please let me know how to do this please as i will be very grateful!!

    thanks!
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  2. #2
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    Quote Originally Posted by andyboy179 View Post
    hi, i need to try and find the angle of VAC and find the height of this pyramid!

    Click image for larger version. 

Name:	pyramid.jpg 
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    to find the height of it would i do:
    1/3x20x15= 100??

    as for finding the angle of VAC i don't know what to do!

    if you can help please let me know how to do this please as i will be very grateful!!

    thanks!
    As far as I understand your sketch ...

    1. The vertex of the pyramid (Obviously you labeled it V) is perpendicularly above the intersection M of the two diagonals of the base area.
    The diagonal has a length of 25.

    2. The height h, a half diagonal and an edge form a right triangle:

    24^2 = 12.5^2+h^2

    Solve for h.

    3. The angle \angle(MVC) is half as large as the angle \angle(AVC). Use the right traingle MCV to determine the angle \angle(MVC):

    \sin(\angle(MVC)) = \dfrac {12.5}{24}~\implies~\angle(MVC)\approx 31.388^\circ
    Attached Thumbnails Attached Thumbnails finding height and angle?-rechteckpyramide.png  
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  3. #3
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    Height

    you said 24^2= 12.5^2 + H^2

    would h= 419.75?

    angle of VAC

    i don't really understand where MVC came from??
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  4. #4
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    Quote Originally Posted by andyboy179 View Post
    Height

    you said 24^2= 12.5^2 + H^2

    would h= 419.75?
    No. That's h^2.

    angle of VAC

    i don't really understand where MVC came from??
    I've attached a new sketch with all the necessary labels so you can see which angles are meant.
    Attached Thumbnails Attached Thumbnails finding height and angle?-rechteckpyramide.png  
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  5. #5
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    Quote Originally Posted by earboth View Post
    No. That's h^2.



    I've attached a new sketch with all the necessary labels so you can see which angles are meant.
    to get the height would i square root 419.75= 20.49??

    i know what the angles are now but how would this help me find the angle of VAC?
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  6. #6
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    Once you know angle VAC, use the fact that the angles in a triangle sum to 180 degrees (or \pi radians) and that, since traingle AVC is an isosceleles triangle, that angles VAC and VCA are equal.
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  7. #7
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    i don't understand how to work all of this out though! its really confusing
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  8. #8
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    Quote Originally Posted by andyboy179 View Post
    i don't understand how to work all of this out though! its really confusing
    1. You should have seen that |\angle(CAV)| = |\angle(VCM)|.

    2. Let d denote the length of one diagonal of the base area and e the length of one edge, then you see that

    \cos(\angle(VCM)) = \dfrac{\frac12 d}{e}.

    3. Calculate the value of \angle(VCM) and subsequently the value of \angle(CAV).
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  9. #9
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    Quote Originally Posted by andyboy179 View Post
    i don't understand how to work all of this out though! its really confusing
    Remember that the law of sines is your friend. I'm sure you can use it in the picture earboth has drawn in order to first calculate the angle MCV, as the angle VMC is 90 degrees. After that you don't need to do much more, as the angle VAC is equal to the angle MCV.
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  10. #10
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    Would VCM=0.3?
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  11. #11
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    Quote Originally Posted by andyboy179 View Post
    Would VCM=0.3?
    No.

    How did you get this result?
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  12. #12
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    Sorry, misread my calculator. My answer was 74.91, to get that I did 12.5x1/2 /24=0.26 inverse cos=74.91??
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  13. #13
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    Quote Originally Posted by earboth View Post
    1. You should have seen that |\angle(CAV)| = |\angle(VCM)|.

    2. Let d denote the length of one diagonal of the base area and e the length of one edge, then you see that

    \cos(\angle(VCM)) = \dfrac{\frac12 d}{e}.

    3. Calculate the value of \angle(VCM) and subsequently the value of \angle(CAV).
    Quote Originally Posted by andyboy179 View Post
    Sorry, misread my calculator. My answer was 74.91, to get that I did 12.5x1/2 /24=0.26 inverse cos=74.91??
    As I've suggested in the post #8 you'll get the Cosine of \angle(VCM) by calculating
    \cos(\angle(VCM)) = \dfrac{\frac12 d}{e}
    \cos(\angle(VCM)) = \dfrac{12.5}{24} = \dfrac{25}{48}

    And now use the inverse Cosine function of your calculator.

    The final result should be \angle(VCM) \approx 58.61^\circ
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  14. #14
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    Why does it go from 12.5/24 to 25/48??
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  15. #15
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    12.5/24=(12.5*2)/(24*2)=25/48
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