krit.club logo

Trigonometry - Trigonometry in 3D

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

3D Pythagoras' Theorem: To find the distance between opposite corners of a cuboid (the space diagonal), we extend the 2D theorem to three dimensions: d2=l2+w2+h2d^2 = l^2 + w^2 + h^2. This is equivalent to finding the diagonal of the base first, then using that as a side in a second right-angled triangle with the height.

A 3D cuboid showing the space diagonal 'd' from a bottom corner to the opposite top corner.
•

The Angle between a Line and a Plane: To find the angle between a line (like a sloped edge) and a plane (like a base), project the line onto the plane. The angle is formed between the original line and its projection on the plane. This always creates a right-angled triangle where the height is the perpendicular distance from the top point to the plane.

Diagram showing a line meeting a plane, its projection on the plane, and the angle between them.
•

Standard Right-Angled Trigonometry in 3D: Once a 3D problem is broken down into 2D right-angled triangles, use SOH CAH TOA: sin⁡θ=OH\sin \theta = \frac{O}{H}, cos⁡θ=AH\cos \theta = \frac{A}{H}, and tan⁡θ=OA\tan \theta = \frac{O}{A}. Identifying the correct right angle is the most critical step.

•

Angle between two Planes: This is the angle between two lines, one in each plane, that meet at right angles to the line of intersection of the two planes.

📐Formulae

d2=x2+y2+z2d^2 = x^2 + y^2 + z^2 (Pythagoras' Theorem in 3D)

sin⁡θ=OppositeHypotenuse\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}

cos⁡θ=AdjacentHypotenuse\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}

tan⁡θ=OppositeAdjacent\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}

Area of a triangle=12absin⁡C\text{Area of a triangle} = \frac{1}{2}ab \sin C

💡Examples

Problem 1:

A cuboid has dimensions AB=8AB = 8 cm, BC=6BC = 6 cm, and height CG=5CG = 5 cm. Calculate the length of the space diagonal AGAG.

Solution:

AG=82+62+52=64+36+25=125≈11.18AG = \sqrt{8^2 + 6^2 + 5^2} = \sqrt{64 + 36 + 25} = \sqrt{125} \approx 11.18 cm.

Explanation:

To find the diagonal of a cuboid, apply the 3D version of Pythagoras' Theorem using the length, width, and height.

Problem 2:

A square-based pyramid has a base side of 10 cm and a vertical height of 12 cm. Find the angle between a sloped edge and the base.

Solution:

  1. Find half the diagonal of the base: Diagonal AC=102+102=200=102AC = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2}. Half diagonal AM=52≈7.071AM = 5\sqrt{2} \approx 7.071 cm.
  2. In the right-angled triangle formed by the height (h=12h=12) and AMAM: tan⁡θ=1252\tan \theta = \frac{12}{5\sqrt{2}}.
  3. θ=tan⁡−1(127.071)≈59.5∘\theta = \tan^{-1}(\frac{12}{7.071}) \approx 59.5^\circ.

Explanation:

The angle between an edge and the base is found by creating a right-angled triangle using the vertical height of the pyramid and the distance from the center of the base to a corner.

Problem 3:

In a cuboid where AB=10AB=10 cm, BC=4BC=4 cm, and height AE=3AE=3 cm, find the angle the diagonal BHBH makes with the base ABCDABCD. (Assume HH is above DD).

Solution:

  1. Find the length of the base diagonal BD=102+42=116≈10.77BD = \sqrt{10^2 + 4^2} = \sqrt{116} \approx 10.77 cm.
  2. The height HD=3HD = 3 cm.
  3. In △BDH\triangle BDH: tan⁡(∠HBD)=HDBD=310.77\tan(\angle HBD) = \frac{HD}{BD} = \frac{3}{10.77}.
  4. ∠HBD=tan⁡−1(0.2785)≈15.6∘\angle HBD = \tan^{-1}(0.2785) \approx 15.6^\circ.

Explanation:

The angle between a line (BH) and a plane (the base) is the angle between the line and its projection on that plane (BD).

Problem 4:

A right pyramid has a rectangular base ABCDABCD with AB=12AB = 12 cm and BC=8BC = 8 cm. The vertex VV is directly above the center MM of the base. The vertical height VMVM is 15 cm. Calculate the angle between the edge VAVA and the base ABCDABCD.

A rectangular-based pyramid with vertex V and height VM.

Solution:

  1. Find the distance AMAM. ACAC is the diagonal of the base: AC=122+82=144+64=208=14.42AC = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} = 14.42 cm.
  2. MM is the midpoint of ACAC, so AM=12×14.42=7.21AM = \frac{1}{2} \times 14.42 = 7.21 cm.
  3. In right-angled triangle VMAVMA, ∠VAM\angle VAM is the required angle.
  4. tan⁡(∠VAM)=VMAM=157.21\tan(\angle VAM) = \frac{VM}{AM} = \frac{15}{7.21}.
  5. ∠VAM=tan⁡−1(157.21)≈64.3∘\angle VAM = \tan^{-1}(\frac{15}{7.21}) \approx 64.3^{\circ}.

Explanation:

To find the angle between a sloped edge and the base, we use the right-angled triangle formed by the edge (hypotenuse), the vertical height, and the distance from the vertex's projection to the corner.

Problem 5:

A wedge-shaped block has a horizontal rectangular base PQRSPQRS where PQ=20PQ = 20 cm and QR=10QR = 10 cm. The vertical face PQUTPQUT is a rectangle with height PT=6PT = 6 cm. Calculate the angle between the plane SURTSURT and the base PQRSPQRS.

A 3D wedge showing a rectangular base and a sloped face.

Solution:

  1. The angle between the planes is the angle between two lines perpendicular to the intersection SRSR. The line PSPS is in the base and PS⊥SRPS \perp SR. The line TSTS is in the sloped plane and TS⊥SRTS \perp SR (as PQUTPQUT is a rectangle and PTPT is vertical).
  2. The required angle is ∠TSP\angle TSP in the right-angled triangle TPSTPS.
  3. PT=6PT = 6 cm (Opposite) and PS=QR=10PS = QR = 10 cm (Adjacent).
  4. tan⁡(∠TSP)=610=0.6\tan(\angle TSP) = \frac{6}{10} = 0.6.
  5. ∠TSP=tan⁡−1(0.6)≈31.0∘\angle TSP = \tan^{-1}(0.6) \approx 31.0^{\circ}.

Explanation:

To find the angle between two planes, identify the line of intersection (SRSR) and find two lines meeting it at right angles (PSPS and TSTS).