CIE IGCSE Mathematics (0580) Pythagoras’ theorem and Trigonometry in 3D Study Notes - New Syllabus
CIE IGCSE Mathematics (0580) Pythagoras’ theorem and Trigonometry in 3D Study Notes
LEARNING OBJECTIVE
- Applying Pythagoras’ Theorem & Trigonometry in 3D
Key Concepts:
- Applying Pythagoras’ Theorem in 3D
- Trigonometry in 3D
- Angle Between a Line and a Plane
Applying Pythagoras’ Theorem in 3D
1. Applying Pythagoras’ Theorem in 3D
In 3D geometry, Pythagoras’ Theorem can be used to calculate lengths such as space diagonals or edges in cuboids, pyramids, and other solids by applying the theorem in two stages.
The basic form in 3D for a cuboid is:
\( \text{Length of space diagonal } d = \sqrt{l^2 + w^2 + h^2} \)
This can be derived in two steps:
- First, find the diagonal of the base using: \( \sqrt{l^2 + w^2} \)
- Then apply Pythagoras again with the height to find the full space diagonal.
Example:
A cuboid has length = 6 cm, width = 8 cm, and height = 5 cm. Find the length of the space diagonal from one corner to the opposite corner.
▶️ Answer/Explanation
Step 1: Find the diagonal of the base:
\( \text{Base diagonal} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{cm} \)
Step 2: Use this with height in 3D Pythagoras:
\( \text{Space diagonal} = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} \approx \boxed{11.18 \, \text{cm}} \)
Example:
In a square-based pyramid, the base has side 10 cm, and the vertical height from the apex to the centre of the base is 12 cm. Find the slant edge from the apex to a corner of the base.
▶️ Answer/Explanation
Step 1: Distance from centre of base to corner:
\( = \frac{1}{2} \cdot \text{diagonal of base} = \frac{1}{2} \cdot \sqrt{10^2 + 10^2} = \frac{1}{2} \cdot \sqrt{200} = \frac{1}{2} \cdot 14.14 = 7.07 \, \text{cm} \)
Step 2: Use Pythagoras with vertical height:
\( \text{Slant edge} = \sqrt{12^2 + 7.07^2} = \sqrt{144 + 50} = \sqrt{194} \approx \boxed{13.93 \, \text{cm}} \)
2. Trigonometry in 3D – Using Sine, Cosine, and Tangent
In 3D geometry, trigonometric ratios (sine, cosine, tangent) are used to find unknown angles or lengths once a right-angled triangle is identified within the 3D figure.
Standard trigonometric ratios:
- \( \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} \)
- \( \cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} \)
- \( \tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} \)
To apply these in 3D:
- Identify a right-angled triangle that contains the known and required quantities.
- Project 3D lengths onto horizontal or vertical planes to simplify into 2D right-angled triangles.
Example:
The shape $ABCDEFGH$ is a cuboid.
$AB = 6\,\text{cm},\quad BG = 3\,\text{cm},\quad FG = 2\,\text{cm}$
$AF = 7\,\text{cm}$
Calculate the angle between $AF$ and the plane $ABCD$.
▶️ Answer/Explanation
Construct right triangle $\triangle AFC$:
$AF = 7\,\text{cm}$ (hypotenuse)
$FC = 3\,\text{cm}$ (perpendicular to base)
Let angle between $AF$ and base $ABCD$ be $x$
Using:
$
\sin x = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{FC}{AF} = \frac{3}{7}
$
$
\sin x \approx 0.428571
$
$
x = \sin^{-1}(0.428571) \approx \boxed{25.4^\circ}
\quad \text{(3 significant figures)}
$
Example:
In a cuboid, point \( A \) is the bottom front-left corner and point \( B \) is the top back-right corner. The cuboid measures 4 m (length), 3 m (width), and 2 m (height). Find the angle between the line \( AB \) and the horizontal base.
▶️ Answer/Explanation
Step 1: Find the diagonal of the base:
\( \text{Base diagonal} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m} \)
Step 2: Use the right triangle formed by vertical height and base diagonal:
\( \tan \theta = \dfrac{2}{5} \Rightarrow \theta = \tan^{-1}\left(\dfrac{2}{5}\right) \approx \boxed{21.8^\circ} \)
3. Angle Between a Line and a Plane
The angle between a line and a plane is defined as the angle between the line and its projection onto the plane. This always forms a right-angled triangle where the vertical or slant edge is the line, and the horizontal leg is its projection.
To calculate this angle:
- Find the horizontal projection of the line onto the plane (often a base diagonal).
- Use trigonometric ratios with the vertical and horizontal parts.
- Use the formula: \( \theta = \tan^{-1}\left(\dfrac{\text{vertical height}}{\text{horizontal projection}}\right) \)
Example:
A triangular prism has sides of 5 cm and 7 cm and is 10 cm long.
What, to the nearest degree, does the line EB make with the base ABCD of the prism?
▶️ Answer/Explanation
To find the angle between line EB and the base of the prism (ABCD), we form triangle DEB.
Step 1: Calculate the length of diagonal DB using Pythagoras’ Theorem:
\( AB^2 = 7^2 + 10^2 = 49 + 100 = 149 \)
\( AB = \sqrt{149} \approx 12.2066 \, \text{cm} \)
Step 2: Use the tangent ratio in triangle DEB (right triangle):
\( \tan(x) = \frac{\text{opposite}}{\text{adjacent}} = \frac{DE}{DB} = \frac{5}{12.2066} \approx 0.4099 \)
\( x = \tan^{-1}(0.4099) \approx 22.27^\circ \)
Final Answer:
\( \boxed{22^\circ} \) (to the nearest degree)
Example:
ABCDEFGH is a cuboid. Work out the angle between the plane DCGH and the line HB.
▶️ Answer/Explanation
We consider triangle HCG in the vertical plane DCGH:
\( HC = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}~\text{cm} \)
Next, consider triangle HBC, where:
- \( HC = 4\sqrt{13} \, \text{cm} \) is adjacent to angle \( \theta \)
- \( BC = 6 \, \text{cm} \) is opposite to angle \( \theta \)
- \( \theta \) is the angle between line HB and the plane DCGH
Use the tangent ratio:
\( \tan(\theta) = \frac{6}{4\sqrt{13}} \)
\( \theta = \tan^{-1} \left( \frac{6}{4\sqrt{13}} \right) \approx \boxed{22.6^\circ} \)