POP

E6.6 Pythagoras' Theorem and Trigonometry in 3D

Image · 914 × 270

01 — Explanation

How do I find a length hidden inside a 3D solid?

Pythagoras and SOHCAHTOA still work in three dimensions — the real skill is spotting the flat right-angled triangle buried inside the solid. Almost every 3D problem is solved by working through two right-angled triangles in sequence: a base triangle first, then a vertical triangle that reuses that result.

02 — Explanation

How do I find the space diagonal of a cuboid?

For a cuboid with length $l$ , width $w$ , height $h$ : first find the diagonal across the base, then combine it with the height.

d = \sqrt{l^2+w^2} \qquad D = \sqrt{d^2+h^2}

For $l=8$ , $w=6$ , $h=5$ cm:

d = \sqrt{8^2+6^2}=\sqrt{100}=10 \text{ cm} \qquad D=\sqrt{10^2+5^2}=\sqrt{125}=11.2 \text{ cm } (3\text{ s.f.})

Examiner Tip

Never substitute a single edge where a diagonal is needed — the angle between a space diagonal and the base always uses the base diagonal, not one wall of the cuboid, as the adjacent side.

03 — Explanation

How do I find the angle between a diagonal and the base?

Using the same cuboid, the triangle formed by the height, the base diagonal, and the space diagonal (hypotenuse) gives:

\tan\theta = \frac{\text{height}}{\text{base diagonal}} = \frac{5}{10} = 0.5 \implies \theta = \tan^{-1}(0.5) = 26.6° \ (1\text{ d.p.})

04 — Explanation

How is the angle between a diagonal and a face different from the angle with the base?

The base is only one of six faces — a diagonal can also be compared with a side face. The working triangle changes accordingly: instead of using the base diagonal, it uses one full edge of the cuboid as the "adjacent" side, since that edge lies within the face being compared.

Example: using the same cuboid ( $l=8$ , $w=6$ , $h=5$ ), find the angle between the space diagonal and the face measuring $8\times5$ .

The diagonal of that face is $\sqrt{8^2+5^2}=\sqrt{89}$ , but the angle with the face is found using the edge perpendicular to that face ( $w=6$ ) as the opposite side and the face's own diagonal as the adjacent side:

\sin\theta = \frac{6}{D} = \frac{6}{11.2} \implies \theta = \sin^{-1}(0.53554...) = 32.4° \ (1\text{ d.p.})

Always identify which face or line a question refers to before drawing the triangle — "the base" and "a side face" give genuinely different right-angled triangles, not just different numbers in the same one.

05 — Explanation

How does this apply to a pyramid?

A square-based pyramid $ABCD$ with vertex $V$ directly above the centre $M$ , side $10$ cm, height $VM=12$ cm — find the angle between edge $VA$ and the base.

Diagonal of the base: $AC = \sqrt{10^2+10^2}=\sqrt{200}$ cm.
$M$ is the midpoint of that diagonal: $AM = \dfrac{\sqrt{200}}{2}=\sqrt{50}$ cm.
In right triangle $VMA$ , the required angle is at $A$ :

\tan\theta = \frac{VM}{AM} = \frac{12}{\sqrt{50}} \implies \theta = \tan^{-1}(1.69706...) = 59.5° \ (1\text{ d.p.})

Carry $\sqrt{50}$ through unrounded rather than converting it to a decimal early — rounding an intermediate length before the final trig step is the most common source of an inaccurate answer here.

06 — Explanation

How do I find the angle between two slant edges of a pyramid?

Using the same pyramid, the angle $\angle VAC$ between edge $VA$ and the base diagonal $AC$ can be found once $VA$ itself is known, since triangle $VMA$ is right-angled at $M$ :

VA = \sqrt{VM^2+AM^2} = \sqrt{12^2+(\sqrt{50})^2} = \sqrt{144+50} = \sqrt{194} = 13.93 \text{ cm}

The angle between two edges that meet at a vertex (like $VA$ and $VC$ ) instead requires the isosceles triangle $VAC$ , using $AC=\sqrt{200}$ as its base and the now-known $VA=VC=\sqrt{194}$ as its equal sides — splitting that isosceles triangle down its line of symmetry converts it into two more right-angled triangles, exactly as in a 2D isosceles problem.