Prove that the angle between the surface of an inclined plane and a the horizontal (the angle of inclination) is equal to the angle between the normal of the plane and the vertical.

Prove that the angle between the surface of an inclined plane and a the horizontal (the angle of inclination) is equal to the angle between the normal of the plane and the vertical.

The angle between the surface of an inclined plane and a the horizontal (the angle of inclination) is equal to the angle between the normal of the plane and the vertical.

Consider a horizontal plane ' A B ' and ' B D ' be its normal. 

Let ' A C ' be the inclined plane and Ө is its angle with the horizontal plane ' A B' and ‘ E C ’ is the normal on plane ' A C ' . 

We have to find out the angle between the normal of the plane and the vertical.

A triangle has three corners , called vertices. The sides of a triangle ( line segments ) that come together at a vertex form two angles ( four angles if you consider the sides of the triangle to be lines instead of line segments ) . Only one of these angles contains the third side of the triangle in its interior, and this angle is called an interior angle of the triangle. In the picture above, the angles ∠ ABC , ∠ BCA and ∠ CAB are the three interior angles of the triangle. An exterior angle is formed by extending one of the sides of the triangle; the angle between the extended side and the other side is the exterior angle. In the diagram , angle ∠ ACD is an exterior angle.

According to exterior angle theorem, the exterior angle at a vertex of a triangle equals the sum of the sizes of the interior angles at the other two vertices of the triangle ( remote interior angles ) . So, in the above diagram , the size of ∠ ACD equals the size of ∠ ABC plus the size of ∠ CAB.

Mathematically it can be written as ,

∠ ACD = ∠ ABC + ∠ CAB                                                                               - - - - ( 1 )

But according to the geometry ,

∠ ACD = ∠ ACE + ∠ ECD                                                                               - - - - ( 2 )

From equation ( 1 ) and ( 2 )

∠ ABC + ∠ CAB = ∠ ACE + ∠ ECD                                                               - - - - ( 3 )

As segment ‘ B D ’ is normal to the horizontal plane ‘ A B ’

∠ ABD = ∠ ABC = 90^O                                                                                                  - - - - ( 4 )

And as ' AC ' is the inclined plane and Ө is its angle with the horizontal plane ' AB'

∠ CAB = Ө                                                                                                       - - - - ( 5 )

As ‘ E C ’ is the normal on plane ' A C '  

∠ ACE = 90^O                                                                                                    - - - - ( 6 )

By using equation ( 4 ) , ( 5 ) & ( 6 ) equation ( 3 ) becomes

90^O  + Ө = 90^O + ∠ ECD

Therefore ,

∠ECD = Ө

i.e.  the angle between the surface of an inclined plane and the horizontal ( the angle of inclination ) is equal to the angle between the normal of the plane and the vertical.