i have a project i am working ion but i am stuck on a question.
i have done parts 1 and 2 i am including them so you can help me with part 3.
1) draw a 3D cube with side length a
2)draw in the diagonal of the cube, from the same corner also draw in the diagonal of the base of the cube.
3) Let Theta be the angle between the Two diagonals, Determine the exact value of theta.
Plz help i am so confused. This is as far as i have gotten and i dont even know if am going the right direction. i have this equation Tan(theta)=a/2a
i have done parts 1 and 2 i am including them so you can help me with part 3.
1) draw a 3D cube with side length a
2)draw in the diagonal of the cube, from the same corner also draw in the diagonal of the base of the cube.
3) Let Theta be the angle between the Two diagonals, Determine the exact value of theta.
Plz help i am so confused. This is as far as i have gotten and i dont even know if am going the right direction. i have this equation Tan(theta)=a/2a
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Well, you are almost right. But the length of the diagonal in the base is not 2a, it is a*sqrt(2).
Use the Pythagorean theorem to determine that the length of the diagonal in the base is a*sqrt(2):
Call the length of the base diagonal d.
d^2 = a^2 + a^2
d^2 = 2*a^2
d = a* sqrt(2))
Use that value in the equation of the tangent:
Tan(theta) = a/(a*sqrt(2)) = 1/sqrt(2) = sqrt(2)/2
Theta = Arctan(sqrt(2)/2)
Theta is approximately. 35.2644 degrees.
Well, you are almost right. But the length of the diagonal in the base is not 2a, it is a*sqrt(2).
Use the Pythagorean theorem to determine that the length of the diagonal in the base is a*sqrt(2):
Call the length of the base diagonal d.
d^2 = a^2 + a^2
d^2 = 2*a^2
d = a* sqrt(2))
Use that value in the equation of the tangent:
Tan(theta) = a/(a*sqrt(2)) = 1/sqrt(2) = sqrt(2)/2
Theta = Arctan(sqrt(2)/2)
Theta is approximately. 35.2644 degrees.
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The exact answer is Arctan(sqrt(2)/2).
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