3D Trigonometry · Study Notes · Edexcel IGCSE 4MA1 Higher

Find lengths and angles in 3D shapes. Identify right triangles within the figure; apply Pythagoras and trig.

At a glance

Identify a RIGHT TRIANGLE within the 3D figure.
Use Pythagoras for unknown lengths.
Use trigonometry (SOH CAH TOA) for angles.
Cuboid diagonals combine 2D and 3D Pythagoras.
Pyramid problems often use HALF the base diagonal.
Angle between line and plane: angle to the projection.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.11 — Apply Pythagoras and trigonometry in 3D contexts.
4.11 — Find angles between lines and planes.
4.11 — Solve cuboid, pyramid, and cone 3D problems.

Method for 3D problems

▼

Find right triangle. Pythagoras for length, trig for angle.

General approach:

Sketch the 3D figure clearly.
Identify the right triangle within the 3D space.
Use Pythagoras to find any unknown sides.
Use SOH CAH TOA for the angle.

Cuboid example. $4 \times 6 \times 12$ . Angle between long diagonal and base.

Right triangle: vertical $= 12$ , floor diagonal $= \sqrt{4^{2} + 6^{2}} = \sqrt{52}$ .
$\tan \theta = 12/\sqrt{52}$ .
$\theta \approx 59.0°$ .

The right triangle: floor diagonal and vertical edge give angle θ to the base.

Pyramid example. Square base side $6$ , perpendicular height $4$ . Slant edge to base.

Half-diagonal of base: $\sqrt{3^{2} + 3^{2}} = \sqrt{18} = 3\sqrt{2}$ .
Slant edge length: $\sqrt{(3\sqrt{2})^{2} + 4^{2}} = \sqrt{18 + 16} = \sqrt{34}$ .
Angle: $\tan \theta = 4 / (3\sqrt{2}) \approx 0.943$ , $\theta \approx 43.3°$ .

Cone example. $r = 5$ , $h = 12$ . Slant length and angle.

Slant: $\sqrt{r^{2} + h^{2}} = \sqrt{169} = 13$ .
Angle to base: $\tan \theta = 12/5 = 2.4$ , $\theta \approx 67.4°$ .

Worked qualitative. Why must we identify a right triangle first?

Trigonometry's basic tools (SOH CAH TOA) only work in right triangles.
3D shapes contain many implicit right triangles.
Drawing them out explicitly helps avoid errors.

Edexcel tip. ALWAYS sketch the right triangle within the 3D figure. Mark schemes credit recognising and labelling.

Find right triangle.
Pythagoras for sides.
Trig for angle.
Sketch is essential.

Common 3D problems

▼

Cuboids, pyramids, cones — standard methods.

Cuboid: long diagonal.

For sides $a, b, c$ :

Long diagonal: $d = \sqrt{a^{2} + b^{2} + c^{2}}$ .
Angle to base: $\tan \theta = c / \sqrt{a^{2} + b^{2}}$ (vertical / floor diagonal).

Pyramid: slant edge angle.

Square-base pyramid with side $s$ and height $h$ :

Half-diagonal of base: $\dfrac{s}{2}\sqrt{2}$ .
Slant edge: $\sqrt{(s/2)^{2} \cdot 2 + h^{2}} = \sqrt{s^{2}/2 + h^{2}}$ .
Angle of slant edge with base: $\tan \theta = \dfrac{h}{(s/2)\sqrt{2}}$ .

The slant edge reaches a base corner — use the half-diagonal, not the full diagonal.

Cone: slant length.

Cone with radius $r$ , height $h$ :

Slant: $l = \sqrt{r^{2} + h^{2}}$ .
Angle to base: $\tan \theta = h/r$ .

Worked qualitative. Why is the half-diagonal used for pyramids, not the full diagonal?

The slant edge goes from a corner of the base to the apex (above the centre).
The horizontal distance from centre to corner = HALF the diagonal.
That's the correct adjacent side.

Edexcel tip. Identify whether the question asks about slant EDGE (corner to apex) or slant FACE (face triangle). They use different distances.

Cuboid: 3D Pythagoras for diagonal.
Pyramid: half-diagonal.
Cone: $r, h, l$ form right triangle.
Sketch carefully.

Quick recap

Find right triangle in 3D.
Pythagoras for unknown side.
Trig for angle.
Cuboid: vertical edge + floor diagonal.
Pyramid: half-diagonal of base.
Cone: $r, h, l$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Cuboid diagonal: $\sqrt{a^{2} + b^{2} + c^{2}}$ .
Floor diagonal of cuboid: $\sqrt{a^{2} + b^{2}}$ .
Cone slant: $l = \sqrt{r^{2} + h^{2}}$ .
Pyramid half-diagonal: $(s/2)\sqrt{2}$ .

How it’s examined

3D trigonometry on Higher Tier (5-7 marks). Often combined with cones, pyramids, or cuboids. Examiner reports flag (1) not finding right triangle, (2) using full diagonal instead of half, (3) wrong angle (slant edge vs face).

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H Jan 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on 3d trigonometry, written exactly the way a tutor would explain them at the board.

1Angle between long diagonal and base
Higher• 3D, cuboid
▼
Question
Cuboid $4 \times 6 \times 12$ . Find the angle between the long diagonal and the base.
Step-by-step solution
1. Step 1
  Floor diagonal: $d_{\text{floor}}^{2} = 16 + 36 = 52$ . $d_{\text{floor}} = \sqrt{52}$ .
2. Step 2
  Right triangle in the diagonal plane: opposite (height) $= 12$ , adjacent (floor diagonal) $= \sqrt{52}$ .
3. Step 3
  $\tan \theta = \dfrac{12}{\sqrt{52}}$ .
4. Step 4
  $\theta = \tan^{-1}(12/\sqrt{52}) \approx 59.0°$ .
Answer
$\theta \approx 59.0°$
Examiner tip
Method: identify the right triangle in 3D. Use Pythagoras for one leg, then trig for the angle.
2Angle in a pyramid
Higher• Adapted from 4MA1/2H Jan 2024 Q19• pyramid
▼
Question
Square-based pyramid: base side $6$ cm, perpendicular height $4$ cm. Find the angle between a slant edge and the base.
Step-by-step solution
1. Step 1
  Base diagonal: $\sqrt{6^{2} + 6^{2}} = \sqrt{72} = 6\sqrt{2}$ . Half-diagonal: $3\sqrt{2}$ .
2. Step 2
  Right triangle: opposite (height) = $4$ , adjacent (half-diagonal) = $3\sqrt{2}$ .
3. Step 3
  $\tan \theta = \dfrac{4}{3\sqrt{2}} \approx 0.943$ .
4. Step 4
  $\theta \approx 43.3°$ .
Answer
$\theta \approx 43.3°$
3Angle in a cone
Higher• cone
▼
Question
Cone with radius $5$ cm and perpendicular height $12$ cm. Find the angle between the slant and the base.
Step-by-step solution
1. Step 1
  Right triangle: opposite (height) $= 12$ , adjacent (radius) $= 5$ .
2. Step 2
  $\tan \theta = 12/5 = 2.4$ .
3. Step 3
  $\theta = \tan^{-1}(2.4) \approx 67.4°$ .
Answer
$\approx 67.4°$
4Naming the angle for a line from a top corner [1 mark]
Getting started• Paper 1H/2H style — 1 mark• 3d trigonometry, line and plane, cuboid
▼
Question
$ABCDEFGH$ is a cuboid. $ABCD$ is the horizontal base, and $E$ , $F$ , $G$ and $H$ are vertically above $A$ , $B$ , $C$ and $D$ respectively.

$M$ is the midpoint of the edge $BC$ .

Write down the name of the angle between the line $EM$ and the plane $ABCD$ . [1]
Step-by-step solution
1. Step 1
  $E$ is vertically above $A$ , so the foot of the perpendicular from $E$ to the base is $A$ . The projection of $EM$ onto the base is $AM$ .
2. Step 2
  B1 The angle is between $EM$ and $AM$ , at $M$ , where the line meets the plane.
  $\text{angle } EMA$
Answer
Angle $EMA$ (or angle $AME$ )
Examiner tip
Examiner tip: The projection does not have to be a diagonal. Here it is the line from $A$ to the midpoint of $BC$ , because $A$ is the point directly below $E$ .

Common mistake: Writing angle $EMB$ . $B$ is not below $E$ , so $MB$ is not the projection and triangle $EMB$ has no right angle at $B$ .
5Light at the centre of a ceiling [2 marks]
Getting started• Paper 1H/2H style — 2 marks• 3d trigonometry, pythagoras, cuboid
▼
Question
A room is a cuboid $5.2$ m long, $4.1$ m wide and $2.6$ m high.

A light is fixed at the centre of the ceiling.

Calculate the straight-line distance from the light to one corner of the floor. Give your answer correct to 3 significant figures. [2]
Step-by-step solution
1. Step 1
  M1 The point on the floor directly below the light is the centre of the floor. From a corner, it is half the length along and half the width across: $2.6$ m and $2.05$ m. The light is then $2.6$ m above that point, so use Pythagoras in 3D with these three distances.
  $d^{2} = 2.6^{2} + 2.05^{2} + 2.6^{2} = 6.76 + 4.2025 + 6.76 = 17.7225$
2. Step 2
  A1 Square root and round.
  $d = \sqrt{17.7225} = 4.2098\ldots = 4.21 \text{ m}$
Answer
$4.21$ m
Examiner tip
Examiner tip: Pythagoras in 3D uses three distances at right angles to each other, not necessarily the edges of the cuboid. Here two of them are half-edges, because the light is at the centre of the ceiling.

Common mistake: Using the full length and width, $\sqrt{5.2^{2} + 4.1^{2} + 2.6^{2}} = 7.11$ m. That is the distance from a floor corner to the opposite ceiling corner, not to the centre of the ceiling.
6Cone from its slant height and radius [2 marks]
Getting started• Paper 1H/2H style — 2 marks• 3d trigonometry, cone, line and plane
▼
Question
A cone has base radius $8$ cm and slant height $20$ cm.

Calculate the size of the angle between the slant edge of the cone and its base. Give your answer correct to 1 decimal place. [2]
Step-by-step solution
1. Step 1
  M1 The vertical cross-section through the vertex gives a right-angled triangle: the radius ( $8$ ) along the base, the height up the axis and the slant height ( $20$ ) as the hypotenuse. The angle at the rim has the radius adjacent to it.
  $\cos\theta = \dfrac{8}{20}$
2. Step 2
  A1 Use the inverse cosine.
  $\theta = 66.421\ldots = 66.4^{\circ}$
Answer
$66.4^{\circ}$
Examiner tip
Examiner tip: Label the sides of the triangle relative to the angle at the rim before choosing the ratio. The slant height is always the hypotenuse.

Common mistake: Using $\tan\theta = \dfrac{8}{20}$ , which gives $21.8^{\circ}$ . The slant height is the hypotenuse, not the opposite side, so tangent is the wrong ratio.
7Pyramid: from sloping edge to slant height [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• 3d trigonometry, pyramid, line and plane
▼
Question
$VABCD$ is a pyramid with a square base $ABCD$ of side $12$ cm. $V$ is vertically above the centre $O$ of the base, and each sloping edge is $16$ cm long.

$M$ is the midpoint of $AB$ .

Calculate the size of the angle between $VM$ and the base $ABCD$ . Give your answer correct to 1 decimal place. [3]
Step-by-step solution
1. Step 1
  M1 Triangle $VAB$ is isosceles, so $VM$ is perpendicular to $AB$ and $AM = 6$ cm. Find $VM$ by Pythagoras.
  $VM^{2} = 16^{2} - 6^{2} = 220 \qquad VM = \sqrt{220} = 14.832\ldots$
2. Step 2
  M1 The projection of $VM$ is $OM$ , which is half the side, $6$ cm. In triangle $VMO$ (right angle at $O$ ), $VM$ is the hypotenuse.
  $\cos VMO = \dfrac{6}{\sqrt{220}}$
3. Step 3
  A1 Use the inverse cosine.
  $VMO = 66.138\ldots = 66.1^{\circ}$
Answer
$66.1^{\circ}$
Examiner tip
Examiner tip: Two different $6$ cm lengths appear here: $AM$ in the sloping face and $OM$ in the base. Label each triangle so you know which one you are using.

Common mistake: Taking $16$ cm as the hypotenuse of triangle $VMO$ and writing $\cos^{-1}(6 \div 16) = 68.0^{\circ}$ . The $16$ cm edge is $VA$ , not $VM$ .
8Pole at the centre of a square courtyard [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• 3d trigonometry, angle of elevation, square
▼
Question
$ABCD$ is a horizontal square courtyard of side $30$ m. A vertical pole $PQ$ stands with its foot $Q$ at the centre of the courtyard.

The angle of elevation of the top of the pole, $P$ , from the corner $A$ is $25^{\circ}$ .

Calculate the height of the pole. Give your answer correct to 3 significant figures. [3]
Step-by-step solution
1. Step 1
  M1 $Q$ is the centre of the square, so $AQ$ is half of the diagonal $AC$ .
  $AC = \sqrt{30^{2} + 30^{2}} = \sqrt{1800} \qquad AQ = \tfrac{1}{2}\sqrt{1800} = 21.213\ldots \text{ m}$
2. Step 2
  M1 The angle of elevation is angle $PAQ$ , in triangle $PAQ$ with the right angle at $Q$ .
  $PQ = 21.213\ldots \times \tan 25^{\circ}$
3. Step 3
  A1 Evaluate and round.
  $PQ = 9.891\ldots = 9.89 \text{ m}$
Answer
$9.89$ m
Examiner tip
Examiner tip: The angle of elevation from a point on the ground is the angle between the line of sight and the ground, so it is found in the vertical triangle through that point and the foot of the pole.

Common mistake: Using half the side, $15$ m, as the distance from $A$ to $Q$ , which gives $6.99$ m. $Q$ is at the centre, so the distance from a corner is half the diagonal.
9Cone: line to the midpoint of a chord [4 marks]
Stretch• Paper 1H/2H style — 4 marks• 3d trigonometry, cone, line and plane
▼
Question
A cone has vertex $V$ and a horizontal circular base with centre $O$ and radius $9$ cm. $V$ is vertically above $O$ and $VO = 15$ cm.

$A$ and $B$ are points on the circumference of the base with $AB = 12$ cm. $M$ is the midpoint of $AB$ .

Calculate the size of the angle between $VM$ and the base. Give your answer correct to 1 decimal place. [4]
Step-by-step solution
1. Step 1
  M1 The line from the centre of a circle to the midpoint of a chord is perpendicular to the chord. So triangle $OMA$ has a right angle at $M$ , with $OA = 9$ and $AM = 6$ .
  $OM^{2} = 9^{2} - 6^{2} = 45$
2. Step 2
  A1 Square root, keeping the exact value.
  $OM = \sqrt{45} = 6.708\ldots \text{ cm}$
3. Step 3
  M1 $V$ is above $O$ , so the projection of $VM$ is $OM$ and the angle is $VMO$ , with the right angle at $O$ .
  $\tan VMO = \dfrac{15}{\sqrt{45}}$
4. Step 4
  A1 Use the inverse tangent.
  $VMO = 65.905\ldots = 65.9^{\circ}$
Answer
$65.9^{\circ}$
Examiner tip
Examiner tip: Give the circle fact as a reason when you use it: "the perpendicular from the centre to a chord bisects the chord". It tells the examiner why triangle $OMA$ is right-angled.

Common mistake: Using the radius, $9$ cm, as $OM$ and giving $\tan^{-1}(15 \div 9) = 59.0^{\circ}$ . That is the angle for a line from $V$ to a point on the circumference, not to the midpoint of a chord. Also, $OM^{2} = 9^{2} + 6^{2}$ adds squares where a shorter side is needed.
10Two angles of elevation of one mast [5 marks]
Stretch• Paper 1H/2H style — 5 marks• 3d trigonometry, angle of elevation, pythagoras
▼
Question
$TM$ is a vertical mast, with its foot $M$ on horizontal ground.

$A$ is a point on the ground due south of $M$ , with $AM = 40$ m. The angle of elevation of $T$ from $A$ is $30^{\circ}$ .

$B$ is a point on the ground due east of $M$ . The angle of elevation of $T$ from $B$ is $22^{\circ}$ .

Calculate the distance $AB$ . Give your answer correct to 3 significant figures. [5]
Step-by-step solution
1. Step 1
  M1 Triangle $TAM$ has a right angle at $M$ . Find the height of the mast.
  $TM = 40 \tan 30^{\circ}$
2. Step 2
  A1 Keep the full value.
  $TM = 23.094\ldots \text{ m}$
3. Step 3
  M1 Triangle $TBM$ has a right angle at $M$ . $TM$ is opposite the $22^{\circ}$ angle and $BM$ is adjacent.
  $BM = \dfrac{23.094\ldots}{\tan 22^{\circ}} = 57.159\ldots \text{ m}$
4. Step 4
  M1 South and east are at right angles, so angle $AMB = 90^{\circ}$ on the ground. Use Pythagoras in triangle $AMB$ .
  $AB^{2} = 40^{2} + 57.159\ldots^{2}$
5. Step 5
  A1 Square root and round.
  $AB = 69.765\ldots = 69.8 \text{ m}$
Answer
$69.8$ m
Examiner tip
Examiner tip: There are three triangles here: two vertical ones through the mast and one flat one on the ground. Sketch each one flat before you calculate.

Common mistake: Adding the two ground distances, $40 + 57.2 = 97.2$ m, as if $A$ , $M$ and $B$ were in a straight line. The directions south and east make a right angle at $M$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
$VABCD$ is a pyramid with a rectangular base $ABCD$ . $O$ is the centre of the base and $V$ is vertically above $O$ .

Write down the name of the angle between the edge $VB$ and the plane $ABCD$ . [1]
Model answer
Projection of $VB$ is $OB$ .

Answer: angle $VBO$
Why this scores
Mark scheme:
- B1 for angle $VBO$ or angle $OBV$ oe
Why this scores full marks: The candidate uses the point directly below $V$ , which is $O$ , and puts $B$ , where the edge meets the base, in the middle. Angle $VBA$ or angle $VBC$ uses an edge of the base instead of the projection and scores B0.
Question 2
Paper 1H/2H style — 2 marks2 marks
$ABCDEFGH$ is a cube of side $9$ cm. $A$ and $G$ are opposite vertices.

Show that $AG = 9\sqrt{3}$ cm. Show your working clearly. [2]
Model answer
$AG^{2} = 9^{2} + 9^{2} + 9^{2} = 243$

$AG = \sqrt{243} = \sqrt{81 \times 3} = \sqrt{81} \times \sqrt{3} = 9\sqrt{3}$
Why this scores
Mark scheme:
- M1 for $9^{2} + 9^{2} + 9^{2}$ oe
- A1 for $\sqrt{243} = \sqrt{81 \times 3}$ leading to $9\sqrt{3}$ , fully shown
Why this scores full marks: The surd is simplified by showing the square factor $81$ . A decimal such as $15.59$ compared with $9\sqrt{3}$ on a calculator does not show the result and scores M1 A0.
Question 3
Paper 1H/2H style — 2 marks2 marks
$ABCDEFGH$ is a cuboid. $ABCD$ is the horizontal base, and $E$ , $F$ , $G$ and $H$ are vertically above $A$ , $B$ , $C$ and $D$ respectively.

$AB = 9$ cm, $BC = 4$ cm and $CG = 6$ cm.

Calculate the size of the angle between the line $BG$ and the plane $ABCD$ . Give your answer correct to 1 decimal place. [2]
Model answer
$G$ is above $C$ , so the projection of $BG$ is $BC$ .

$\tan GBC = \dfrac{6}{4} \qquad GBC = 56.30\ldots$

Answer: $56.3^{\circ}$
Why this scores
Mark scheme:
- M1 for $\tan GBC = \dfrac{6}{4}$ oe
- A1 for awrt $56.3$
Why this scores full marks: $BG$ lies in the vertical face $BCGF$ , so its projection is simply the edge $BC$ and no Pythagoras is needed. Using the base diagonal $\sqrt{9^{2} + 4^{2}}$ as the adjacent side gives $31.4^{\circ}$ and scores M0.
Question 4
Paper 1H/2H style — 3 marks3 marks
$ABCDEFGH$ is a cuboid. The base $ABCD$ is horizontal with $AB = 8$ cm and $BC = 6$ cm, and $G$ is vertically above $C$ .

The angle between the diagonal $AG$ and the plane $ABCD$ is $40^{\circ}$ .

Calculate the length of $AG$ . Give your answer correct to 3 significant figures. [3]
Model answer
$AC = \sqrt{8^{2} + 6^{2}} = 10$

Angle $GAC = 40^{\circ}$ , right angle at $C$ .

$\cos 40^{\circ} = \dfrac{10}{AG} \qquad AG = \dfrac{10}{\cos 40^{\circ}} = 13.054\ldots$

Answer: $13.1$ cm
Why this scores
Mark scheme:
- M1 for $AC = 10$
- M1 for $\cos 40 = \dfrac{10}{AG}$ oe
- A1 for awrt $13.1$
Why this scores full marks: $AC$ is adjacent to the $40^{\circ}$ angle and $AG$ is the hypotenuse, so cosine is the right ratio. Using sine gives $10 \div \sin 40^{\circ} = 15.6$ cm and loses the last two marks.
Question 5
Paper 1H/2H style — 3 marks3 marks
A cone has perpendicular height $16$ cm. The angle between the slant edge and the base is $62^{\circ}$ .

Calculate the diameter of the base of the cone. Give your answer correct to 3 significant figures. [3]
Model answer
Right-angled triangle: height $16$ opposite the $62^{\circ}$ angle, radius $r$ adjacent.

$\tan 62^{\circ} = \dfrac{16}{r} \qquad r = \dfrac{16}{\tan 62^{\circ}} = 8.507\ldots$

$\text{diameter} = 2 \times 8.507\ldots = 17.01\ldots$

Answer: $17.0$ cm
Why this scores
Mark scheme:
- M1 for $\tan 62 = \dfrac{16}{r}$ oe
- M1 for $r = \dfrac{16}{\tan 62}$ ( $= 8.507\ldots$ )
- A1 for awrt $17.0$
Why this scores full marks: The candidate answers the question asked. Stopping at the radius, $8.51$ cm, scores M1 M1 A0.
Question 6
Paper 1H/2H style — 4 marks4 marks
$VABCD$ is a pyramid with a square base $ABCD$ . $V$ is vertically above the centre $O$ of the base and $VO = 12$ cm.

The angle between the edge $VA$ and the base is $55^{\circ}$ .

Calculate the length of a side of the base. Give your answer correct to 3 significant figures. [4]
Model answer
Angle $VAO = 55^{\circ}$ , right angle at $O$ .

$AO = \dfrac{12}{\tan 55^{\circ}} = 8.402\ldots$

$AC = 2 \times 8.402\ldots = 16.804\ldots$

$AB^{2} + AB^{2} = AC^{2} \qquad AB = \dfrac{16.804\ldots}{\sqrt{2}} = 11.882\ldots$

Answer: $11.9$ cm
Why this scores
Mark scheme:
- M1 for $\tan 55 = \dfrac{12}{AO}$ oe
- A1 for $AO = 8.40$ or better
- M1 for $AB = \dfrac{2 \times \text{their } AO}{\sqrt{2}}$ oe
- A1 for awrt $11.9$
Why this scores full marks: The script works backwards through the same chain as a forward question: angle to $AO$ , $AO$ to the diagonal, diagonal to the side. Giving $AO = 8.40$ cm as the side of the base scores only the first two marks.
Question 7
Paper 1H/2H style — 4 marks4 marks
The curved surface area of a cone is $150\pi$ cm². The radius of the base is $6$ cm.

Calculate the size of the angle between the slant edge and the base. Give your answer correct to 1 decimal place. [4]
Model answer
$\pi r l = 150\pi \qquad 6l = 150 \qquad l = 25$

$\cos\theta = \dfrac{6}{25} \qquad \theta = 76.11\ldots$

Answer: $76.1^{\circ}$
Why this scores
Mark scheme:
- M1 for $\pi \times 6 \times l = 150\pi$ oe
- A1 for $l = 25$
- M1 for $\cos\theta = \dfrac{6}{\text{their } l}$ oe
- A1 for awrt $76.1$
Why this scores full marks: The curved surface area formula $\pi r l$ is on the Higher formula sheet, and the candidate divides both sides by $\pi$ instead of evaluating it, so $l$ comes out exact. Using $\sin\theta = \dfrac{6}{25}$ gives $13.9^{\circ}$ , the angle between the slant edge and the vertical axis.
Question 8
Paper 1H/2H style — 5 marks5 marks
$ABCD$ is a triangular-based pyramid. The base $ABC$ is horizontal with $AB = AC = 10$ cm and angle $BAC = 40^{\circ}$ . $D$ is vertically above $A$ and $AD = 7$ cm. $M$ is the midpoint of $BC$ .

(a) Calculate the length of $BC$ . Give your answer correct to 3 significant figures. [2]

(b) Calculate the size of the angle between $DM$ and the plane $ABC$ . Give your answer correct to 1 decimal place. [3]
Model answer
(a)

$BC^{2} = 10^{2} + 10^{2} - 2 \times 10 \times 10 \times \cos 40^{\circ} = 46.79\ldots$

$BC = 6.840\ldots$

Answer: $6.84$ cm

(b) Triangle $ABC$ is isosceles, so $AM$ is perpendicular to $BC$ and $BM = 3.420\ldots$

$AM^{2} = 10^{2} - 3.420\ldots^{2} \qquad AM = 9.396\ldots$

$D$ is above $A$ , so the angle is $DMA$ , right angle at $A$ .

$\tan DMA = \dfrac{7}{9.396\ldots} \qquad DMA = 36.68\ldots$

Answer: $36.7^{\circ}$
Why this scores
Mark scheme:
- M1 for $10^{2} + 10^{2} - 2 \times 10 \times 10 \times \cos 40$
- A1 for awrt $6.84$
- M1 for $AM = \sqrt{10^{2} - (\tfrac{1}{2} \times \text{their } BC)^{2}}$ or $AM = 10\cos 20$
- M1 for $\tan DMA = \dfrac{7}{\text{their } AM}$ oe
- A1 for awrt $36.7$
Why this scores full marks: The cosine rule is copied correctly from the formula sheet, which the June 2024 examiner report noted some candidates failed to do. In (b) the candidate states why $AM$ is perpendicular to $BC$ before using a right angle at $M$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
$ABCD$ is a rectangular ramp. The edge $AB$ lies on horizontal ground and $AB = 8$ m. The edges $AD$ and $BC$ are each $5$ m long and slope upwards at $20^{\circ}$ to the horizontal, so that $DC$ is horizontal.

Calculate the size of the angle between the diagonal $AC$ of the ramp and the horizontal ground. Give your answer correct to 1 decimal place. [5]
Model answer
Height of $C$ above the ground:

$h = 5 \sin 20^{\circ} = 1.710\ldots$

$ABCD$ is a rectangle, so

$AC = \sqrt{8^{2} + 5^{2}} = \sqrt{89} = 9.433\ldots$

Let $X$ be the point on the ground directly below $C$ . Triangle $CAX$ has a right angle at $X$ and hypotenuse $AC$ .

$\sin CAX = \dfrac{1.710\ldots}{\sqrt{89}} \qquad CAX = 10.44\ldots$

Answer: $10.4^{\circ}$
Why this scores
Mark scheme:
- M1 for $5 \sin 20$
- A1 for $1.71$ or better
- M1 for $AC = \sqrt{8^{2} + 5^{2}}$ oe
- M1 for $\sin CAX = \dfrac{\text{their } h}{\text{their } AC}$ oe
- A1 for awrt $10.4$
Why this scores full marks: The candidate introduces the point $X$ below $C$ , so the right-angled triangle is named. The height and the length $AC$ are both measured in 3D, and sine links them directly. Using $\tan^{-1}(1.71 \div 8) = 12.1^{\circ}$ treats $AB$ as the projection of $AC$ , which it is not.
Question 10
Paper 1H/2H style — 6 marks6 marks
$VABCD$ is a pyramid with a square base $ABCD$ of side $15$ cm. $V$ is vertically above the centre $O$ of the base. $M$ is the midpoint of $AB$ .

The volume of the pyramid is $1050$ cm³.

Volume of a pyramid $= \tfrac{1}{3} \times \text{area of base} \times \text{height}$

(a) Calculate the size of the angle between $VM$ and the base. Give your answer correct to 1 decimal place. [4]

(b) Calculate the size of the angle between $VA$ and the base. Give your answer correct to 1 decimal place. [2]
Model answer
(a)

$\tfrac{1}{3} \times 15^{2} \times h = 1050 \qquad 75h = 1050 \qquad h = 14$

$OM = 7.5$ cm, right angle at $O$ .

$\tan VMO = \dfrac{14}{7.5} \qquad VMO = 61.82\ldots$

Answer: $61.8^{\circ}$

(b)

$AO = \tfrac{1}{2}\sqrt{15^{2} + 15^{2}} = 10.606\ldots$

$\tan VAO = \dfrac{14}{10.606\ldots} \qquad VAO = 52.85\ldots$

Answer: $52.9^{\circ}$
Why this scores
Mark scheme:
- M1 for $\tfrac{1}{3} \times 15^{2} \times h = 1050$ oe
- A1 for $h = 14$
- M1 for $\tan VMO = \dfrac{\text{their } h}{7.5}$ oe
- A1 for awrt $61.8$
- M1 for $\tan VAO = \dfrac{\text{their } h}{\tfrac{1}{2}\sqrt{15^{2} + 15^{2}}}$ oe
- A1 for awrt $52.9$
Why this scores full marks: The candidate uses the two different projections correctly: $OM$ is half a side for the line to the midpoint of an edge, and $AO$ is half a diagonal for the sloping edge. Using the full diagonal in (b) gives $33.4^{\circ}$ and scores M0 A0.

4MA1 3D Trigonometry

At a glance

What you’ll learn

Method for 3D problems

Common 3D problems

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Angle between long diagonal and base

2Angle in a pyramid

3Angle in a cone

4Naming the angle for a line from a top corner [1 mark]

5Light at the centre of a ceiling [2 marks]

6Cone from its slant height and radius [2 marks]

7Pyramid: from sloping edge to slant height [3 marks]

8Pole at the centre of a square courtyard [3 marks]

9Cone: line to the midpoint of a chord [4 marks]

10Two angles of elevation of one mast [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10