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Trigonometry - Trigonometry in 3D

Grade 10IGCSE

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

🔑Concepts

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

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Applying Pythagoras' Theorem in three dimensions to find the distance between two opposite vertices.

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Projecting a line onto a plane to find the angle between a line and a plane.

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Visualizing the 'slanted height' versus 'vertical height' in pyramids and cones.

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Using trigonometric ratios (SOH CAH TOA) on 2D planes extracted from 3D figures.

📐Formulae

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

sin⁡heta=OppositeHypotenuse\sin heta = \frac{\text{Opposite}}{\text{Hypotenuse}}

cos⁡heta=AdjacentHypotenuse\cos heta = \frac{\text{Adjacent}}{\text{Hypotenuse}}

tan⁡heta=OppositeAdjacent\tan heta = \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).