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Trigonometry - 3D Trigonometry

Grade 11IGCSE

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

🔑Concepts

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Identifying 2D right-angled triangles within 3D structures (cuboids, pyramids, cones).

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Calculating the space diagonal of a cuboid using 3D Pythagoras.

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Finding the angle between a line and a plane by identifying the projection of the line onto that plane.

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Applying the Sine Rule and Cosine Rule to non-right-angled triangles formed by internal slices of 3D shapes.

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Calculating the angle between two faces (dihedral angle) by finding the line of greatest slope perpendicular to the intersection line.

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Using angles of elevation and depression within a three-dimensional context.

📐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.