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Trigonometry - 3D Trigonometry

Grade 11A Level

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

🔑Concepts

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3D Trigonometry involves solving problems in three-dimensional shapes like cuboids, pyramids, and prisms by identifying right-angled triangles embedded within them. The first step is often to use the Pythagorean theorem in 2D to find a diagonal on the base before moving to a vertical triangle.

A cuboid showing a base diagonal and a space diagonal forming internal triangles.
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The angle between a line and a plane is the angle between the line and its projection on that plane. For example, the angle between space diagonal AGAG and base ABCDABCD is ∠GAC\angle GAC.

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For pyramids, the height always meets the base at its center (centroid). For a square-based pyramid, the height drops to the intersection of the diagonals of the square base.

A pyramid showing the vertical height meeting the center of the base.
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The angle between two planes (dihedral angle) is found by drawing two lines, one in each plane, that both meet the line of intersection at 90∘90^\circ at the same point.

📐Formulae

Pythagoras in 3D: d2=x2+y2+z2d^2 = x^2 + y^2 + z^2

Sine Rule: asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Cosine Rule (Length): a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A

Cosine Rule (Angle): cos⁡A=b2+c2−a22bc\cos A = \frac{b^2 + c^2 - a^2}{2bc}

Basic Trig: sin⁡θ=OH,cos⁡θ=AH,tan⁡θ=OA\sin \theta = \frac{O}{H}, \cos \theta = \frac{A}{H}, \tan \theta = \frac{O}{A}

Area of a triangle: Area=12absin⁡C\text{Area} = \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 and the angle AGAG makes with the base ABCDABCD.

Solution:

  1. Find diagonal of the base ACAC: AC=82+62=64+36=10AC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = 10 cm.
  2. Find AGAG using △ACG\triangle ACG: AG=AC2+CG2=102+52=125≈11.18AG = \sqrt{AC^2 + CG^2} = \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.18 cm.
  3. Find angle θ=∠GAC\theta = \angle GAC: tan⁡θ=CGAC=510=0.5\tan \theta = \frac{CG}{AC} = \frac{5}{10} = 0.5.
  4. θ=tan⁡−1(0.5)≈26.6∘\theta = \tan^{-1}(0.5) \approx 26.6^{\circ}.

Explanation:

To find the space diagonal, we first apply Pythagoras to the horizontal base to find ACAC. Then, we use the vertical triangle ACGACG where ACAC is the base and CGCG is the height. The angle between the line AGAG and the base is the angle between the line and its projection ACAC on the base.

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 sloping face and the base.

Solution:

  1. Let MM be the midpoint of one base edge and OO be the center of the base.
  2. The distance OM=12×10=5OM = \frac{1}{2} \times 10 = 5 cm.
  3. The vertical height VO=12VO = 12 cm.
  4. In the right-angled △VOM\triangle VOM, let the angle at MM be α\alpha.
  5. tan⁡α=VOOM=125=2.4\tan \alpha = \frac{VO}{OM} = \frac{12}{5} = 2.4.
  6. α=tan⁡−1(2.4)≈67.4∘\alpha = \tan^{-1}(2.4) \approx 67.4^{\circ}.

Explanation:

The angle between a sloping face and the base is measured along the line of greatest slope. We create a right-angled triangle using the vertical height of the pyramid, the distance from the center of the base to the midpoint of the edge, and the slant height of the face.

Problem 3:

A triangular prism has a horizontal rectangular base ABCDABCD where AB=15AB = 15 cm and BC=8BC = 8 cm. The vertical face ABFEABFE is a rectangle with height AE=6AE = 6 cm. Calculate the length of the diagonal ECEC and the angle it makes with the base ABCDABCD.

Diagram of a triangular prism with base ABCD and vertical height AE.

Solution:

  1. Find ACAC (diagonal of the base) using Pythagoras: AC2=AB2+BC2=152+82=225+64=289AC^2 = AB^2 + BC^2 = 15^2 + 8^2 = 225 + 64 = 289 AC=289=17 cmAC = \sqrt{289} = 17\text{ cm}

  2. Use △EAC\triangle EAC (right-angled at AA) to find ECEC: EC2=AE2+AC2=62+172=36+289=325EC^2 = AE^2 + AC^2 = 6^2 + 17^2 = 36 + 289 = 325 EC=325≈18.03 cmEC = \sqrt{325} \approx 18.03\text{ cm}

  3. Find the angle θ=∠ECA\theta = \angle ECA: tan⁡θ=AEAC=617\tan \theta = \frac{AE}{AC} = \frac{6}{17} θ=arctan⁡(617)≈19.4∘\theta = \arctan\left(\frac{6}{17}\right) \approx 19.4^\circ

Explanation:

We first calculate the diagonal of the base to create a right-angled triangle EACEAC that contains the space diagonal and the angle required.

Problem 4:

A right pyramid has a square base of side 12 cm. The sloping edges are all 10 cm long. Calculate the vertical height of the pyramid.

Square-based pyramid with labels for height VM and slant edge VA.

Solution:

  1. Let the base be ABCDABCD with center MM and vertex VV. Diagonal ACAC of the square base: AC=122+122=144+144=288=122≈16.97 cmAC = \sqrt{12^2 + 12^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \approx 16.97\text{ cm}

  2. The distance from a corner to the center MM is half the diagonal: AM=12AC=62≈8.485 cmAM = \frac{1}{2} AC = 6\sqrt{2} \approx 8.485\text{ cm}

  3. In △VMA\triangle VMA (right-angled at MM): VM2+AM2=VA2VM^2 + AM^2 = VA^2 h2+(62)2=102h^2 + (6\sqrt{2})^2 = 10^2 h2+72=100h^2 + 72 = 100 h2=28h^2 = 28 h=28=27≈5.29 cmh = \sqrt{28} = 2\sqrt{7} \approx 5.29\text{ cm}

Explanation:

To find the vertical height, we construct a right-angled triangle using the slant edge (hypotenuse) and the distance from the vertex center to the corner.