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Thread: Right triangular pyramid

  1. #1
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    Right triangular pyramid

    I have a right triangular pyramid, and I need to calculate an angle at its apex.
    (more precisely, the angles between the edges that join the apex to the vertices at the base that define the hypotenuse of the base)
    (ie the angle that spans the hypotenuse of the base)

    The apex is directly above the right angle of the base.
    Therefore, all three of the angles at the vertix below the apex are right angles.

    I know the other two angles at the apex, I just need the angle that spans the hypotenuse of the base.
    It should be fairly simple, but my calculations fall short. More importantly, whilst I will summarise my efforts below, I feel there should be a simpler way of calculating this angle.

    I initially tried assigning an arbritrary value (x, say) to the length between the apex and the right angle.
    This length forms a shared side of two right angle triangles. Using SOHCAHTOA, I calculated all the lengths of these triangles.
    I then used pythagoras' theorem to calculate the hypotenuse of the base.
    Finally, looking at the triangle formed by the three hypotenuses, I used the cosine rule to calculate the angle spanning the base hypotenuse.

    I think my logic is sound but doubtless I've made some mistake somewhere. More importantly, as I say, surely there's a simpler formula for calculating this angle based purely on the other two angles of the apex.
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  2. #2
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    Re: Right triangular pyramid

    let your base be $ABC$ with apex $D$

    $\angle ABC$ is a right angle, point $A$ is $(0,y)$, point $B$ is $(0,0)$, point $C$ is $(x,0)$, point $D$ is $(0,0,h)$

    This should describe your pyramid.

    I believe you want the angles, $\angle CBD$, $\angle BAD$

    $\angle CBD =\arccos\left(\dfrac{BC \cdot CD}{|BC||CD|}\right) = \arccos\left(\dfrac{-x}{\sqrt{x^2+h^2}}\right)$

    $\angle BAD = \arccos\left(\dfrac{BA \cdot AD}{|BA||AD|}\right) = \arccos\left(\dfrac{-y}{\sqrt{y^2+h^2}}\right)$
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  3. #3
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    Re: Right triangular pyramid

    Thanks for your answer. That does describe my pyramid.
    The angle I want is $\angle ADC$

    The known angles are $\angle ADB$ and $\angle BDC$

    Can your forumla be rearranged:


    $\angle ADC =\arccos\left(\dfrac{DA \cdot AC}{|DA||AC|}\right)$
    Last edited by Cametre; Dec 22nd 2016 at 06:56 AM.
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  4. #4
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    Re: Right triangular pyramid

    Quote Originally Posted by Cametre View Post
    Thanks for your answer. That does describe my pyramid.
    The angle I want is $\angle ADC$

    The known angles are $\angle ADB$ and $\angle BDC$

    Can your forumla be rearranged:


    $\angle ADC =\arccos\left(\dfrac{DA \cdot AC}{|DA||AC|}\right)$
    yessir
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