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Angle between planes of a pyramid
Find angle between two linesInverse Trigonometry Plots ArcT(T(x))-ClarificationTrig Integration and Completing the SquareWhy do we always obtain expressions independent of trig. functions when we implicitly derive their inverses?Trigonometry-The distance between L and B when L has walked a distance d must be less than kd, where k is some constant. Find k.How to find one leg if only the angles are known in a right angleHow do I determine if measurements given for a triangle will produce one, two, or no triangles?Prove that tan15a- tan10a-tan5a= tan 15a +tan 10a +tan 5asolve equation with cos and powersFinding an angle inside an irregular quadrilateral
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enter image description here
Question d and e, please help
Thank you
trigonometry
asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
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enter image description here
Question d and e, please help
Thank you
trigonometry
asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
1
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Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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What have you tried? What I would do, is to calculate the bisection vectors (perpendicular to the base) and calculate the angle between those vectors.
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– Matti P.
17 hours ago
add a comment |
$begingroup$
enter image description here
Question d and e, please help
Thank you
trigonometry
asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
1
New contributor
Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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enter image description here
Question d and e, please help
Thank you
trigonometry
trigonometry
asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
1
New contributor
Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
1
New contributor
Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
1
New contributor
Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
1
asked 17 hours ago
Nuthanon KittiwatanachodNuthanon Kittiwatanachod
1
1
New contributor
Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nuthanon Kittiwatanachod is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
What have you tried? What I would do, is to calculate the bisection vectors (perpendicular to the base) and calculate the angle between those vectors.
$endgroup$
– Matti P.
17 hours ago
add a comment |
$begingroup$
What have you tried? What I would do, is to calculate the bisection vectors (perpendicular to the base) and calculate the angle between those vectors.
$endgroup$
– Matti P.
17 hours ago
$begingroup$
What have you tried? What I would do, is to calculate the bisection vectors (perpendicular to the base) and calculate the angle between those vectors.
$endgroup$
– Matti P.
17 hours ago
$begingroup$
What have you tried? What I would do, is to calculate the bisection vectors (perpendicular to the base) and calculate the angle between those vectors.
$endgroup$
– Matti P.
17 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To find angle between the planes you need to find angle between two lines, one in each plane, that are perpendicular to line formed by intersecting planes. In case of $VPS$ and $VRQ$ the planes intersection line is not obvious but you have to realize that it will be parallel to to $PS$ and $QR$ (fold a piece of papar in such a way that two opposite edges are parallel).
Now draw a perpendicular from $V$ to $PS$, let's say it intersects $PS$ at $X$ and draw a perpendicular from $V$ to $QR$, let's say it intersects $QR$ at $Y$. If you look at triangle $VXY$, you need to find angle $XVY$. We know that $VX$ and $VY$ are perpendicular to the planes intersection line because they are perpendicular to lines that are parallel to the intersection line.
Using Pythagorean theorem, $VX=VY=sqrt9^2-3^2=sqrt72=6sqrt2$. We know that $XY=6$ so $$cosangleXVY=fracVX^2+VY^2-XY^22cdot VXcdot VY=frac144-36144=frac34$$ Thus $mangleXVY approx 41.4$ degrees. Hopefully you can do $e)$ yourself now.
answered 16 hours ago
VasyaVasya
3,9531618
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$begingroup$
To find angle between the planes you need to find angle between two lines, one in each plane, that are perpendicular to line formed by intersecting planes. In case of $VPS$ and $VRQ$ the planes intersection line is not obvious but you have to realize that it will be parallel to to $PS$ and $QR$ (fold a piece of papar in such a way that two opposite edges are parallel).
Now draw a perpendicular from $V$ to $PS$, let's say it intersects $PS$ at $X$ and draw a perpendicular from $V$ to $QR$, let's say it intersects $QR$ at $Y$. If you look at triangle $VXY$, you need to find angle $XVY$. We know that $VX$ and $VY$ are perpendicular to the planes intersection line because they are perpendicular to lines that are parallel to the intersection line.
Using Pythagorean theorem, $VX=VY=sqrt9^2-3^2=sqrt72=6sqrt2$. We know that $XY=6$ so $$cosangleXVY=fracVX^2+VY^2-XY^22cdot VXcdot VY=frac144-36144=frac34$$ Thus $mangleXVY approx 41.4$ degrees. Hopefully you can do $e)$ yourself now.
answered 16 hours ago
VasyaVasya
3,9531618
$endgroup$
add a comment |
$begingroup$
To find angle between the planes you need to find angle between two lines, one in each plane, that are perpendicular to line formed by intersecting planes. In case of $VPS$ and $VRQ$ the planes intersection line is not obvious but you have to realize that it will be parallel to to $PS$ and $QR$ (fold a piece of papar in such a way that two opposite edges are parallel).
Now draw a perpendicular from $V$ to $PS$, let's say it intersects $PS$ at $X$ and draw a perpendicular from $V$ to $QR$, let's say it intersects $QR$ at $Y$. If you look at triangle $VXY$, you need to find angle $XVY$. We know that $VX$ and $VY$ are perpendicular to the planes intersection line because they are perpendicular to lines that are parallel to the intersection line.
Using Pythagorean theorem, $VX=VY=sqrt9^2-3^2=sqrt72=6sqrt2$. We know that $XY=6$ so $$cosangleXVY=fracVX^2+VY^2-XY^22cdot VXcdot VY=frac144-36144=frac34$$ Thus $mangleXVY approx 41.4$ degrees. Hopefully you can do $e)$ yourself now.
answered 16 hours ago
VasyaVasya
3,9531618
$endgroup$
add a comment |
$begingroup$
To find angle between the planes you need to find angle between two lines, one in each plane, that are perpendicular to line formed by intersecting planes. In case of $VPS$ and $VRQ$ the planes intersection line is not obvious but you have to realize that it will be parallel to to $PS$ and $QR$ (fold a piece of papar in such a way that two opposite edges are parallel).
Now draw a perpendicular from $V$ to $PS$, let's say it intersects $PS$ at $X$ and draw a perpendicular from $V$ to $QR$, let's say it intersects $QR$ at $Y$. If you look at triangle $VXY$, you need to find angle $XVY$. We know that $VX$ and $VY$ are perpendicular to the planes intersection line because they are perpendicular to lines that are parallel to the intersection line.
Using Pythagorean theorem, $VX=VY=sqrt9^2-3^2=sqrt72=6sqrt2$. We know that $XY=6$ so $$cosangleXVY=fracVX^2+VY^2-XY^22cdot VXcdot VY=frac144-36144=frac34$$ Thus $mangleXVY approx 41.4$ degrees. Hopefully you can do $e)$ yourself now.
answered 16 hours ago
VasyaVasya
3,9531618
$endgroup$
To find angle between the planes you need to find angle between two lines, one in each plane, that are perpendicular to line formed by intersecting planes. In case of $VPS$ and $VRQ$ the planes intersection line is not obvious but you have to realize that it will be parallel to to $PS$ and $QR$ (fold a piece of papar in such a way that two opposite edges are parallel).
Now draw a perpendicular from $V$ to $PS$, let's say it intersects $PS$ at $X$ and draw a perpendicular from $V$ to $QR$, let's say it intersects $QR$ at $Y$. If you look at triangle $VXY$, you need to find angle $XVY$. We know that $VX$ and $VY$ are perpendicular to the planes intersection line because they are perpendicular to lines that are parallel to the intersection line.
Using Pythagorean theorem, $VX=VY=sqrt9^2-3^2=sqrt72=6sqrt2$. We know that $XY=6$ so $$cosangleXVY=fracVX^2+VY^2-XY^22cdot VXcdot VY=frac144-36144=frac34$$ Thus $mangleXVY approx 41.4$ degrees. Hopefully you can do $e)$ yourself now.
answered 16 hours ago
VasyaVasya
3,9531618
answered 16 hours ago
VasyaVasya
3,9531618
answered 16 hours ago
VasyaVasya
3,9531618
answered 16 hours ago
VasyaVasya
3,9531618
3,9531618
add a comment |
add a comment |
Nuthanon Kittiwatanachod is a new contributor. Be nice, and check out our Code of Conduct.
Nuthanon Kittiwatanachod is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What have you tried? What I would do, is to calculate the bisection vectors (perpendicular to the base) and calculate the angle between those vectors.
$endgroup$
– Matti P.
17 hours ago