Solve three-dimensional problems involving right-angled triangles, including finding the relevant right-angled triangle inside a solid such as a rectangular prism or pyramid, the angle between a line and a plane, and problems that combine right-triangle results across different planes

What this dot point is asking

Trigonometry and Pythagoras are tools for flat triangles, yet the HSC sets them loose inside solids: the diagonal of a shipping container, the slant edge of a pyramid roof, a guy rope on a tent. Nothing new about the ratios is needed here. The whole skill is finding the one right-angled triangle inside the solid that carries the answer, drawing that triangle on its own, and then using exactly the right-angled triangle trigonometry you already know. Everything on this page is in degrees.

A three-dimensional figure usually hides several right-angled triangles, and the marks are for picking the useful ones and chaining them. The reliable approach has four habits. Sketch the solid clearly, large, in oblique projection. Hunt for the right angles, because a vertical edge meeting a horizontal floor is a right angle, and so is the corner of any rectangular face. Redraw the relevant triangle flat, as a plain two-dimensional triangle with its real side lengths marked, so you stop trying to do trigonometry on a perspective drawing. Work around the figure: often one triangle gives you a length (frequently a base diagonal from Pythagoras), and that length is then a side of a second triangle that gives the angle. Get the right triangle and the chain is short.

The answer

Break the solid into right-angled triangles

A solid is a collection of points, edges and flat faces. Trigonometry only ever acts on a single flat triangle, so the first move is always to isolate a triangle that lies in one plane and is right-angled. Two facts generate almost every such triangle in Year 11 work:

• A vertical line meeting a horizontal plane makes a right angle with every horizontal line through the foot of the vertical. A flagpole, a tent pole or the height of a pyramid is vertical, so it is perpendicular to the floor, and that is your right angle.
• Each face of a rectangular prism is a rectangle, so the two edges meeting at any corner of a face are perpendicular, and a diagonal drawn across a face splits it into two right-angled triangles.

The single most useful triangle is the one that turns a flat (base-plane) Pythagoras result into a vertical problem: compute a length lying in the floor (a base diagonal, say), then stand a vertical height on it and you have a new right-angled triangle whose hypotenuse is the slanting line you actually care about.

The diagonals of a rectangular box

A rectangular prism (box) with edge lengths $l$, $w$ and $h$ has two kinds of diagonal, and both come from the same idea, applied once or twice.

A base diagonal lies in one rectangular face. By Pythagoras in that face, a diagonal of the $l$ by $w$ base has length $\sqrt{l^2 + w^2}$.

The space diagonal runs from one corner of the box to the diametrically opposite corner, passing through the inside. It is the hypotenuse of a right-angled triangle whose two legs are a base diagonal (length $\sqrt{l^2 + w^2}$, lying flat) and a vertical edge (length $h$). The right angle sits where the vertical edge meets the base. So

That tidy formula is just Pythagoras done twice. The diagram below shows the box with this triangle picked out.

The angle the space diagonal makes with the base is the angle in that triangle at the bottom corner, between the base diagonal and the space diagonal. Since the height is opposite it and the base diagonal is adjacent, tangent gives it directly.

The angle between a line and a plane

This is the one genuinely new idea. A line that pierces a plane (without being perpendicular to it) leans over at some angle, and we need a single, well-defined number for "how steep" it is. The definition is built from a shadow.

Drop a perpendicular from a point on the line straight down onto the plane; the foot of that perpendicular, joined back along the plane, traces the projection of the line onto the plane, its shadow if the light came straight down. The angle between the line and the plane is the angle between the line and its projection. It is always the smallest angle the line makes with any line in the plane, and it is acute.

In practice this is easier than it sounds, because the projection is usually an edge or diagonal already in your diagram. For the space diagonal of a box, the shadow on the floor is precisely the base diagonal, so the angle between the space diagonal and the base is the angle between the space diagonal and the base diagonal, which is the triangle angle above. For a pyramid's slant edge, the shadow is the line from the base corner to the centre of the base.

A pyramid: the apex above the base centre

For a right pyramid on a square base, the apex sits directly above the centre of the base, and the vertical height drops from the apex to that centre, meeting the base at a right angle. This single fact unlocks the pyramid, because it ties the apex to the one special point from which the base diagonals are easy.

Let the square base have side $a$ and let the apex be height $H$ above the centre $M$. Two slanting lines matter, and each has its own right-angled triangle through $M$:

• The slant edge runs from the apex down to a base corner. Its shadow on the base is the line from the corner to the centre, which is half a base diagonal, length $\dfrac{\sqrt{a^2 + a^2}}{2} = \dfrac{a\sqrt{2}}{2}$. The slant edge is the hypotenuse of the right-angled triangle with legs $H$ (vertical) and $\dfrac{a\sqrt{2}}{2}$ (the half-diagonal), right angle at $M$.
• A slant face meets the base along a base edge; the line from the apex to the midpoint of that edge has its shadow running from the centre to the same midpoint, length $\dfrac{a}{2}$ (this gives the angle between the face and the base, a closely related idea).

The diagram shows a square pyramid with the slant-edge triangle picked out.

The subtle point students miss is the half-diagonal. The line from a base corner to the centre is half a diagonal, not half a side, so its length is $\dfrac{a\sqrt{2}}{2}$ (which is $\dfrac{a}{\sqrt 2}$), not $\dfrac{a}{2}$. Use $\dfrac{a}{2}$ only for the distance from the centre to the midpoint of an edge, the apothem, which belongs to the slant-face triangle, not the slant-edge triangle.

Combining two-dimensional results across planes

The hardest Year 11 three-dimensional questions never leave the toolkit; they just need two triangles in different planes, solved in order. The pattern is: solve a triangle lying in the base (often by Pythagoras) to get a length you do not start with, then carry that length into a second, vertical triangle to get the angle or height asked for. The elevation-from-two-positions problem is the classic case where both sightings share one vertical plane and the unknowns are found by setting two tangent equations equal. The worked examples below drill exactly these chains.

How exam questions ask about three-dimensional trigonometry

The wording signals which triangle to build:

• "Find the length of the diagonal of the prism / the space diagonal" points to Pythagoras done twice: base diagonal first ($\sqrt{l^2 + w^2}$), then lift it with the height to $\sqrt{l^2 + w^2 + h^2}$.
• "Find the angle the diagonal (or edge) makes with the base / with the plane" is a line-plane angle: identify the projection (the shadow on the base), and use $\tan(\text{angle}) = \dfrac{\text{height}}{\text{projection}}$.
• "In a square pyramid of height ... and base ..." means the apex is above the centre; the half-diagonal $\dfrac{a\sqrt2}{2}$ is the projection for a slant edge, and $\dfrac{a}{2}$ for a slant face.
• "Use triangle $XYZ$ to find ..., hence find ..." is a scaffold telling you the exact chain of triangles; do them in the named order, and the "hence" answer feeds off the first.
• "From two points ... the angle of elevation is ..." is two right-angled triangles in one vertical plane; write a tangent equation from each and solve them together.
• "Show that ... = [given value]" hands you the next part's number; lay the working out so it lands exactly on the stated value.
• "Find the exact length" forbids a decimal; keep surds such as $5\sqrt{3}$ or $8\sqrt{2}$.

Finding the right triangle stage by stage

The single most valuable habit in three dimensions is to lift the right triangle out of the solid, drawing it flat once you have found the lengths it needs. Here is that process for the courtyard flagpole worked above: courtyard $ABCD$ is $24$ m by $18$ m, a vertical pole of height $11$ m stands at corner $A$ with top $T$, and we want the angle of elevation of $T$ from the far corner $C$.

Stage 1, draw the solid and mark what you know. Sketch the rectangular courtyard in oblique projection and stand the vertical pole at $A$. Mark every given length: the floor is $24$ m by $18$ m and the pole is $11$ m.

Stage 2, solve a triangle in the base plane first. The distance from $C$ to the pole is the floor diagonal $AC$, which is not given, so find it by Pythagoras in the floor: $AC = \sqrt{24^2 + 18^2} = \sqrt{900} = 30$ m. The right angle here is at $B$, where the two sides of the rectangle meet.

Stage 3, lift the relevant vertical triangle out. Now stand the pole up over $A$. The triangle that holds the answer is $ACT$: the floor diagonal $AC = 30$ m is one leg, the vertical pole $AT = 11$ m is the other, and they meet at a right angle at $A$. The line of sight $CT$ is the hypotenuse.

Stage 4, solve the flat triangle. With the triangle drawn flat, the elevation $\theta$ at $C$ has the height $11$ opposite it and the diagonal $30$ adjacent, so $\tan\theta = \dfrac{11}{30}$, giving $\theta = \tan^{-1}\!\left(\dfrac{11}{30}\right) \approx 20\degree$.

The two-stage chain (Pythagoras in the base, then tangent in the vertical triangle) is the backbone of nearly every three-dimensional question in this course. Find the base length first, lift the right triangle out, then solve it flat.