Trigonometry in solids

In grade 10 you learnt to solve any triangle, not only a right-angled one:

To find a length or an angle in a solid, find a plane triangle that holds it, work out its sides – usually in right-angled triangles – and solve it.

In a right pyramid $\displaystyle SABCD$ on a square base, all the edges equal 4.
Find the angle between the faces $\displaystyle SAB$ and $\displaystyle SAD$.

The solution below draws perpendiculars to $\displaystyle SA$ in the two faces, and finds the angle between them by the law of cosines.

• The face $\displaystyle SAB$ is an equilateral triangle, so the perpendicular from $\displaystyle B$ to $\displaystyle SA$ goes down to the midpoint $\displaystyle K$ of $\displaystyle SA$, and $\displaystyle {BK}={\frac{\sqrt{3}}{2}\cdot4}={2\sqrt{3}}$.
• The same holds in the face $\displaystyle SAD$: $\displaystyle DK\perp SA$ and $\displaystyle {DK}={2\sqrt{3}}$. So the angle between the faces is $\displaystyle \angle BKD$.
• $\displaystyle BD$ is the square’s diagonal: $\displaystyle {BD}={4\sqrt{2}}$.
• In the triangle $\displaystyle BKD$: $\displaystyle {\cos\angle BKD}={\frac{12+12-32}{2\cdot2\sqrt{3}\cdot2\sqrt{3}}}={-\frac{1}{3}}$, so $\displaystyle \angle BKD\approx109.47^{\circ}$.

A plane at distance $\displaystyle d$ from the centre $\displaystyle M$ of a sphere of radius $\displaystyle R$, where $\displaystyle d<R$, cuts the sphere in a circle.
What is the circle’s radius?

• The perpendicular from $\displaystyle M$ to the plane goes down to the circle’s centre.
• Every point of the circle lies on the sphere, at distance $\displaystyle R$ from $\displaystyle M$.
• By Pythagoras’ theorem: $\displaystyle {r}={\sqrt{R^2-d^2}}$, and the circle’s circumference is $\displaystyle 2\pi\sqrt{R^2-d^2}$.

In the box $\displaystyle ABCDA'B'C'D'$: $\displaystyle {AB}={3}$, $\displaystyle {AD}={4}$, $\displaystyle {AA'}={12}$.
Find the angle $\displaystyle {\alpha}={\angle B'AD'}$, rounded to two decimal places.

All the edges of a triangular pyramid $\displaystyle SABC$ equal 6 (a regular tetrahedron).
The foot of the height, $\displaystyle O$, is where the base’s medians meet, so $\displaystyle AO$ is $\displaystyle \frac{2}{3}$ of the median.

A plane at distance 6 from the centre of a sphere of radius 10 cuts the sphere.
Find the circumference of the circle of intersection, exactly.

A sphere of radius 3 is inscribed in a right cone: it touches the base and the lateral surface.
The cone’s slant height makes an angle of $\displaystyle 60^{\circ}$ with the base.
The cross-section through the cone’s axis is an equilateral triangle $\displaystyle SAB$, and the circle inscribed in it is a cross-section of the sphere.