Today's lesson
Real shapes are 3D — pyramids, prisms, masts, ramps, boxes. The trick is to spot a right-angled triangle hidden inside the 3D object, extract it as a clean 2D diagram, then use the same SOHCAHTOA tools you already know.
Learning intentions
- Visualise and pick out a right-angled triangle inside a 3D object
- Redraw that triangle as a clean 2D diagram with all known sides & angles labelled
- Use trig (and sometimes Pythagoras' theorem first) to find an unknown side or angle
- State the answer back in the language of the original 3D object
Part 1 — How to attack a 3D problem (~6 min)
The 3-step routine
- Find the right triangle inside the 3D object (the triangle that contains the unknown and two pieces you can figure out).
- Redraw that triangle on its own, in 2D, with the right angle marked. Sometimes you need Pythagoras first to get one side (e.g. a face diagonal of a cube).
- Solve with SOHCAHTOA, then translate back: "the angle the cable makes with the ground is …", "the height of the mast is …".
Part 2 — Right triangles in a cube (Building understanding, ~10 min)
📺 Walkthrough: a 2 m cube — extract the right triangle $ACD$ (face-diagonal + vertical edge + space-diagonal) and find $\angle DAC$.
(a) Find the exact length of the face-diagonal $AC$ (use Pythagoras).
(b) Find the exact length of the space-diagonal $AD$.
(c) Use trigonometry to find $\angle DAC$, to 1 d.p.
(d) Find $\angle CAB$.
- (a) Triangle $ABC$ is right-angled at $B$ with $AB=BC=2$. Pythagoras: $AC=\sqrt{2^2+2^2}=\boxed{2\sqrt{2}\text{ m}}\approx 2.83$ m.
- (b) Triangle $ACD$ is right-angled at $C$ (the vertical $CD$ is perpendicular to the floor). $AD=\sqrt{(2\sqrt{2})^2+2^2}=\sqrt{12}=\boxed{2\sqrt{3}\text{ m}}\approx 3.46$ m.
- (c) In triangle $ACD$: $\tan(\angle DAC)=\dfrac{CD}{AC}=\dfrac{2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}$, so $\angle DAC=\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)\approx\boxed{35.3°}$.
- (d) In triangle $ABC$ (the bottom face): both legs are 2, so it's an isoceles right triangle and $\angle CAB=\boxed{45°}$.
Take-away: a 3D problem is just a chain of 2D right triangles — work outward from what you know, often using a result from one triangle as a side of the next.
Part 3 — Vertical mast with two cables (Example 9, ~14 min)
📺 Walkthrough: a vertical mast supported by cable A (36 m at 48°). Find the height of the mast, then use it to find the angle of cable B from the horizontal.
(a) Find the height of the mast (to 3 d.p.).
(b) Find the angle the cable from $B$ makes with the horizontal (to 2 d.p.).
- (a) Extract the right triangle on the $A$-side: hypotenuse $36$, angle $48°$, opposite side $h$ (mast height).
- $\sin 48^\circ=\dfrac{h}{36}\;\Rightarrow\;h=36\sin 48^\circ\approx\boxed{26.753\text{ m}}$.
- (b) Now extract the right triangle on the $B$-side: vertical side is the same mast height $h=26.753$, horizontal $24$, unknown angle $\theta$ at $B$.
- $\tan\theta=\dfrac{26.753}{24}$
- $\theta=\tan^{-1}\!\left(\dfrac{26.753}{24}\right)\approx\boxed{48.11°}$ from the horizontal.
Now you try: Cable from $A$ is $53$ m at $37°$, point $B$ is $29$ m from base. (a) $h=53\sin 37°\approx 31.896$ m; (b) $\theta=\tan^{-1}\!\left(\dfrac{31.896}{29}\right)\approx 47.72°$.
Practice 3.1 — 3D applications
- A rectangular box has dimensions $3\times 4\times 5$ cm.
(i) Find the length of the diagonal of its $3\times 4$ base.
(ii) Find the length of the space diagonal of the box.
(iii) Find the angle the space diagonal makes with the base, to 2 d.p. - A square-based pyramid has base side $6$ m and slant height $5$ m (from apex to the midpoint of a base edge). Find (i) the vertical height of the apex above the base, and (ii) the angle the slant face makes with the base.
- An observer stands $40$ m from the base of a $30$ m building, which has a $12$ m flagpole on top. From the observer's eye level (treat as ground level), find the angle of elevation to (i) the top of the building, (ii) the top of the flagpole. (2 d.p.)
a-ii) space diag = $\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt{2}\approx 7.07$ cm
a-iii) $\theta=\tan^{-1}\!\left(\dfrac{5}{5}\right)=45°$
b-i) vertical height = $\sqrt{5^2-3^2}=4$ m
b-ii) angle = $\tan^{-1}\!\left(\dfrac{4}{3}\right)\approx 53.13°$
c-i) $\tan^{-1}\!\left(\dfrac{30}{40}\right)\approx 36.87°$
c-ii) $\tan^{-1}\!\left(\dfrac{42}{40}\right)\approx 46.40°$
Part 4 — Quick quiz (5 min)
Pick the correct answer for each, then click Mark.
Q1. The first step in a 3D trig problem is:
Q2. A cube has side $a$. The length of one face diagonal is:
Q3. In a cube of side $2$ m, the angle between the space diagonal and the base (face diagonal) is:
Q4. A 36 m cable is at $48°$ to the horizontal. The vertical height it reaches is:
Q5. A square-based pyramid has base side $8$ and slant height $5$ (apex to midpoint of base edge). The vertical height is:
Working program — Cambridge Ex 6E
| Section | Foundation | Standard (set work) | Advanced |
|---|---|---|---|
| Fluency 1–5 | 1–4 | 1, 2, 4, 5 | 3–5 |
| Problem-solving 6–9 | 6, 7 | 6–8 | 7–9 |
| Reasoning 10–11 | 10 | 10 | 10, 11 |
| Enrichment 12–13 | — | — | 12, 13 |
Exit ticket — write in your book
- What's the first thing to do when faced with a 3D trig question?
- For a $4$ m cube, what's the length of one face diagonal? (exact value)
- If a $20$ m cable runs from the ground to the top of a $12$ m mast, what angle does it make with the ground?