A line perpendicular to a plane

In grades 10 and 11 you learnt: in a right-angled triangle, $\displaystyle \alpha$ is an acute angle, $\displaystyle a$ the leg opposite it, $\displaystyle b$ the leg next to it and $\displaystyle c$ the hypotenuse.

• The special angles: $\displaystyle {\sin30^{\circ}}={\frac{1}{2}}$, $\displaystyle {\sin45^{\circ}}={\frac{\sqrt{2}}{2}}$, $\displaystyle {\sin60^{\circ}}={\frac{\sqrt{3}}{2}}$, $\displaystyle {\tan30^{\circ}}={\frac{\sqrt{3}}{3}}$, $\displaystyle {\tan60^{\circ}}={\sqrt{3}}$.
• An angle from a ratio of sides: $\displaystyle {\alpha}={\tan^{-1}\left(\frac{a}{b}\right)}$, and likewise $\displaystyle \sin^{-1}$ and $\displaystyle \cos^{-1}$. The calculator in degrees; a rounded angle is written after $\displaystyle \approx$.
• In an exact answer a root stays a root: $\displaystyle {\sqrt{50}}={5\sqrt{2}}$.

A line that meets a plane at the point $\displaystyle A$ is perpendicular to the plane when it is perpendicular to every line in the plane through $\displaystyle A$.

The sufficient condition: if a line is perpendicular to two intersecting lines of the plane that pass through the point where it meets the plane, it is perpendicular to the plane.

In the box, $\displaystyle AA'\perp AB$ and $\displaystyle AA'\perp AD$: the faces $\displaystyle ABB'A'$ and $\displaystyle ADD'A'$ are rectangles. $\displaystyle AB$ and $\displaystyle AD$ meet at $\displaystyle A$, so $\displaystyle AA'$ is perpendicular to the base plane. It follows that $\displaystyle AA'$ is perpendicular to $\displaystyle AC$ too, though $\displaystyle AC$ is not an edge: the triangle $\displaystyle A'AC$ is right-angled at $\displaystyle A$.

The height of a solid is a perpendicular segment from a part of the solid (its top, a vertex) to the base plane. So the height is perpendicular to every line of the base through its foot, and each such line makes a right-angled triangle with the height.

In the box $\displaystyle ABCDA'B'C'D'$: $\displaystyle {AB}={4}$, $\displaystyle {BC}={3}$, $\displaystyle {AA'}={12}$.
Find the length of the diagonal $\displaystyle A'C$ and the angle between $\displaystyle A'C$ and $\displaystyle AC$.

• In the triangle $\displaystyle ABC$, right-angled at $\displaystyle B$: $\displaystyle {AC}={\sqrt{4^2+3^2}}={5}$.
• $\displaystyle AA'$ is perpendicular to the base plane, so $\displaystyle AA'\perp AC$. In the triangle $\displaystyle A'AC$, right-angled at $\displaystyle A$: $\displaystyle {A'C}={\sqrt{12^2+5^2}}={13}$.
• The angle $\displaystyle {\alpha}={\angle A'CA}$: $\displaystyle {\tan\alpha}={\frac{12}{5}}$, so $\displaystyle {\alpha}={\tan^{-1}\left(\frac{12}{5}\right)\approx67.38^{\circ}}$.

In the box $\displaystyle ABCDA'B'C'D'$: $\displaystyle {AB}={6}$, $\displaystyle {BC}={8}$, $\displaystyle {AA'}={10}$.

Why is the triangle $\displaystyle A'AC$ right-angled? Explain.

In the box $\displaystyle ABCDA'B'C'D'$, the base’s diagonal $\displaystyle BD$ is 5 long.
The angle between the box’s diagonal $\displaystyle BD'$ and the base’s diagonal $\displaystyle BD$ is $\displaystyle 30^{\circ}$.

Why is the triangle $\displaystyle BDD'$ right-angled? Explain.

The edge of the cube $\displaystyle ABCDA'B'C'D'$ is 4 long.
Find the angle $\displaystyle \alpha$ between the cube’s diagonal $\displaystyle AC'$ and the base’s diagonal $\displaystyle AC$. Round to two decimal places.

A carpenter puts up a pole and checks with a set square that it is perpendicular to the ground. He checks twice, in two different directions on the ground. Why is one check not enough?