🧊 2-D and 3-D Problems

OCR FSMQ Additional Maths · Trigonometry (PT6)

Level 3 · Ages 15–16

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1 The Golden Rule of 3-D Trigonometry
PT6: apply Pythagoras' Theorem and trigonometry to 2- and 3-dimensional problems. The specification's own example is "find the angle of greatest slope".
The golden rule
Never do trigonometry on a 3-D picture.
Identify the right triangle, redraw it flat and on its own, label it, then solve.
A 3-D diagram is drawn in perspective, so angles and lengths on the page are distorted. A right angle can look like $60°$. Once you have extracted the relevant triangle onto a fresh 2-D sketch, every ordinary technique works.
Do not round the bridging length. If a face diagonal is $\sqrt{61}$, keep it as $\sqrt{61}$ (or in a calculator memory) for the next step. Rounding to $7.81$ early can move the final angle by a whole degree.
2 Diagonals of a Cuboid
A C G face diagonal AC space diagonal AG Find AC first (2-D Pythagoras on the base), then use triangle ACG
The two-stage Pythagoras
Base diagonal:  $AC^2 = a^2 + b^2$
Space diagonal:  $AG^2 = AC^2 + c^2 = a^2 + b^2 + c^2$
Worked Example 1 — A cuboid's space diagonal and angle

A cuboid measures $6$ cm by $4$ cm by $3$ cm. Find the length of the space diagonal, and the angle it makes with the base.

①Base diagonal: $AC = \sqrt{6^2 + 4^2} = \sqrt{52} = 7.2111$ cm
②Space diagonal: $AG = \sqrt{52 + 3^2} = \sqrt{61} = 7.8102$ cm, so $7.81$ cm (3 s.f.).
③Now redraw triangle $ACG$ flat. It is right-angled at $C$, with base $AC = \sqrt{52}$ and height $CG = 3$.
④$\tan(\angle GAC) = \dfrac{3}{\sqrt{52}} = \dfrac{3}{7.2111} = 0.41603$
⑤$\angle GAC = 22.6°$ (1 d.p.)
The formula $\sqrt{a^2+b^2+c^2}$ gives the space diagonal in one step, but you still need $AC$ separately for the angle — so compute it anyway.
Worked Example 2 — The angle between a diagonal and an edge

For the same cuboid, find the angle between the space diagonal $AG$ and the vertical edge $CG$.

①Same triangle $ACG$, right-angled at $C$, but now the angle at $G$.
②$\tan(\angle AGC) = \dfrac{AC}{CG} = \dfrac{7.2111}{3} = 2.4037$
③$\angle AGC = 67.4°$ (1 d.p.)
④Check: $22.6° + 67.4° = 90°$ ✓ — the two non-right angles of the triangle.
3 Pyramids
The key construction
Drop a perpendicular from the apex to the centre of the base.
That vertical line is the height, and it forms a right angle with every line in the base.
Worked Example 3 — A square-based pyramid

A pyramid has a square base of side $8$ cm and four equal sloping edges of length $10$ cm. Find (a) the vertical height, (b) the angle a sloping edge makes with the base, (c) the angle a sloping face makes with the base.

①(a) Half the base diagonal: the full diagonal is $\sqrt{8^2+8^2} = \sqrt{128} = 11.314$, so half is $5.6569$ cm.
②Redraw the triangle: apex, base centre, base corner. Right-angled at the centre, hypotenuse $10$, base $5.6569$.
③$h = \sqrt{100 - 32} = \sqrt{68} = 8.2462$, so $8.25$ cm (3 s.f.).
④(b) In that same triangle, $\cos\theta = \dfrac{5.6569}{10} = 0.56569$
⑤$\theta = 55.6°$ (1 d.p.)
⑥(c) For a face, use the midpoint of a base edge instead of a corner. Its distance from the centre is half the side, $4$ cm.
⑦$\tan\phi = \dfrac{h}{4} = \dfrac{8.2462}{4} = 2.0616$
⑧$\phi = 64.1°$ (1 d.p.)
Edge angle and face angle are different. The face angle ($64.1°$) is steeper than the edge angle ($55.6°$), because the midpoint of a side is closer to the centre than a corner is. Read the question very carefully.
4 The Angle of Greatest Slope
On a sloping plane, different directions climb at different rates. Walking along a contour is level; walking straight up the fall line is steepest. The angle of greatest slope is the angle of that steepest direction.
How to find it
The line of greatest slope runs perpendicular to the horizontal contour of the plane.
$\tan(\text{greatest slope}) = \dfrac{\text{vertical rise}}{\text{shortest horizontal distance}}$
Worked Example 4 — A ramp

A rectangular ramp $ABCD$ has $AB$ horizontal on the ground, of length $5$ m. The opposite edge $DC$ is also horizontal but raised $2$ m, and the ramp's sloping width (from $AB$ to $DC$) is $6$ m. Find the angle of greatest slope.

①$AB$ and $DC$ are both horizontal, so they are the contours. The line of greatest slope is perpendicular to them — that is, straight across the $6$ m width.
②Redraw the vertical triangle: hypotenuse $6$ m (up the slope), vertical rise $2$ m.
③$\sin\theta = \dfrac{2}{6} = 0.33333$
④$\theta = 19.5°$ (1 d.p.)
⑤The horizontal distance covered is $\sqrt{36-4} = \sqrt{32} = 5.657$ m, and indeed $\tan^{-1}\left(\tfrac{2}{5.657}\right) = 19.5°$ ✓
Walking along $AB$ gives a slope of $0°$ — it is level. Any diagonal route is somewhere between $0°$ and $19.5°$. That is why $19.5°$ is the greatest slope.
Worked Example 5 — A diagonal route

On the same ramp, a person walks from $A$ diagonally to $C$. Find the angle of elevation of this route.

①Horizontally, $A$ to $C$ covers $5$ m along the contour and $5.657$ m across.
②Horizontal distance $= \sqrt{5^2 + 5.657^2} = \sqrt{25 + 32} = \sqrt{57} = 7.550$ m
③The rise is still $2$ m.
④$\tan\theta = \dfrac{2}{7.550} = 0.26489$, so $\theta = 14.8°$ (1 d.p.)
⑤As expected, $14.8° < 19.5°$ — the diagonal is a gentler climb, which is why paths zigzag up hills.
5 Quick Reference

Golden rule

Redraw the triangle flat before calculating.

Bridging length

Usually a face diagonal, by Pythagoras.

Keep it exact

Store surds; do not round mid-problem.

Space diagonal

$\sqrt{a^2+b^2+c^2}$.

Pyramid height

Apex to base centre, perpendicular.

Edge angle

Use half the base diagonal.

Face angle

Use half the base side.

Greatest slope

Perpendicular to the contour.

Diagonal routes

Always gentler than the greatest slope.

Elevation

$\tan\theta = \dfrac{\text{rise}}{\text{horizontal distance}}$.

6 Practice Questions
Question 1

Find the length of the space diagonal of a cube of side $5$ cm.

▶ Show solution

$\sqrt{25+25+25} = \sqrt{75} = 5\sqrt3 = 8.66$ cm (3 s.f.)

Question 2

A cuboid is $9$ cm by $12$ cm by $8$ cm. Find the length of the base diagonal of the $9 \times 12$ face.

▶ Show solution

$\sqrt{81 + 144} = \sqrt{225} = 15$ cm

Question 3

For the cuboid in Question 2, find the space diagonal.

▶ Show solution

$\sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17$ cm

Question 4

For the same cuboid, find the angle the space diagonal makes with the $9 \times 12$ base.

▶ Show solution

In the right-angled triangle with base $15$ and height $8$:

$\tan\theta = \dfrac{8}{15} = 0.53333$

$\theta = 28.1°$ (1 d.p.)

Question 5

Find the angle a space diagonal of a cube makes with the base.

▶ Show solution

Take side $1$. Base diagonal $= \sqrt2$, height $= 1$.

$\tan\theta = \dfrac{1}{\sqrt2} = 0.70711$

$\theta = 35.3°$ (1 d.p.) — the same for every cube, whatever its size.

Question 6

A ramp rises $1.5$ m over a sloping length of $9$ m. Find its angle of slope.

▶ Show solution

$\sin\theta = \dfrac{1.5}{9} = 0.16667$

$\theta = 9.6°$ (1 d.p.)

Question 7

A pyramid has a square base of side $6$ cm and vertical height $4$ cm. Find the length of a sloping edge.

▶ Show solution

Base diagonal $= \sqrt{36+36} = \sqrt{72} = 8.4853$, so half is $4.2426$ cm.

Sloping edge $= \sqrt{4.2426^2 + 4^2} = \sqrt{18 + 16} = \sqrt{34}$

$= 5.83$ cm (3 s.f.)

Question 8

For the pyramid in Question 7, find the angle a sloping face makes with the base.

▶ Show solution

For a face, use half the side: $3$ cm.

$\tan\phi = \dfrac{4}{3} = 1.33333$

$\phi = 53.1°$ (1 d.p.)

Question 9

A vertical flagpole is held by a wire from its top to a point $7$ m from its base. The wire is $12$ m long. Find the height of the pole and the angle the wire makes with the ground.

▶ Show solution

$h = \sqrt{144 - 49} = \sqrt{95} = 9.75$ m (3 s.f.)

$\cos\theta = \dfrac{7}{12} = 0.58333$, so $\theta = 54.3°$ (1 d.p.)

Question 10

A rectangular field $ABCD$ lies on a hillside. $AB = 40$ m runs horizontally along a contour. The edge $BC = 30$ m runs directly up the slope, rising $9$ m vertically from $B$ to $C$.

(a) Find the angle of greatest slope of the field.   (b) Find the horizontal distance from $B$ to $C$.   (c) Find the straight-line distance $AC$.   (d) Find the angle of elevation of $C$ from $A$, and explain why it is smaller than the answer to (a).

▶ Show solution

(a) $BC$ runs directly up the slope, so it is the line of greatest slope.

$\sin\theta = \dfrac{9}{30} = 0.3$

$\theta = 17.5°$ (1 d.p.)

(b) Let $B'$ be the point directly below $C$ at the level of $AB$.

$BB' = \sqrt{30^2 - 9^2} = \sqrt{900 - 81} = \sqrt{819} = 28.618$ m, so $28.6$ m (3 s.f.).

(c) $AB$ is horizontal and perpendicular to the direction of $BB'$ (since $AB$ follows a contour and $BB'$ runs directly away from it).

So the horizontal distance from $A$ to the point below $C$ is

$\sqrt{40^2 + 819} = \sqrt{1600 + 819} = \sqrt{2419} = 49.183$ m

Then $AC = \sqrt{2419 + 9^2} = \sqrt{2500} = 50$ m exactly.

(d) $\tan(\text{elevation}) = \dfrac{9}{49.183} = 0.18299$

Elevation $= 10.4°$ (1 d.p.)

It is smaller because the route from $A$ to $C$ is a diagonal across the slope, not straight up it. The same $9$ m of climb is spread over $49.2$ m of horizontal travel instead of $28.6$ m, so the average gradient is gentler.

Only the direction perpendicular to the contours achieves the greatest slope; every other direction on the plane is less steep, and moving along a contour ($A$ to $B$) is not a climb at all.

2-D and 3-D Problems (PT6) · OCR FSMQ Additional Maths · Created with MathJax