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The two sides of an equilateral triangle form angles equal to alpha with the plane P, and the third side lies in this plane. Find the dihedral angle formed by the plane of the triangle and the plane P if sinα=(√3)/4

Sagot :

sqdancefan

Answer:

  30°

Step-by-step explanation:

You want the dihedral angle formed by an equilateral triangle and plane P, given one leg of the triangle lies in plane P and the other legs form an angle α with the plane such that sin(α) = (√3)/4.

Diagram

The attachment shows a diagram of this geometry. The equilateral triangle is ∆ABC, with segment AB in the z-plane (plane P). We have made AB=BC=AC=1, so sin(α) is the height of point C above plane P, (√3)/4.

As in any equilateral triangle, the altitude of the triangle, OC, is (√3)/2 times the side length. This make ∆OCP a right triangle with hypotenuse (√3)/2 and leg CP = (√3)/4.

Dihedral angle POC is ...

  [tex]\angle POC=\arcsin\left(\dfrac{(\dfrac{\sqrt{3}}{4})}{(\dfrac{\sqrt{3}}{2})}\right)=\arcsin\left(\dfrac{2}{4}\right)=30^\circ[/tex]

The dihedral angle between the plane of the triangle and plane P is 30°.

View image sqdancefan
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