Trigonometry in 3 dimensions - Higher
Click to explore updated revision resources for GCSE Maths: Higher - 2D and 3D trigonometry problems, with step-by-step slideshows, quizzes, practice exam questions, and more!
The trigonometric ratios can be used to solve An object with width, height and depth, eg a cube. problems which involve calculating a length or an angle in a right-angled triangle.
It may be necessary to use Pythagoras' theorem and trigonometry to solve a problem.
Example
The shape ABCDEFGH is a cuboid.
Length AB is 6 cm, length BG is 3 cm and length FG is 2 cm.
The length of the diagonal AF is 7 cm.
Calculate the angle between AF and the base ABCD. Give the answer to 3 significant figures.
ABCD is the base of the cuboid. The line FC and the base ABCD form a right angle.
Draw the right-angled triangle AFC and label the sides. The angle between AF and AC is \(x\).
Use \(\sin{x} = \frac{o}{h}\)
\(\sin{x} = \frac{3}{7}\)
\(\sin{x} = 0.428571 \dotsc\). Do not round this answer yet.
To calculate the angle use the inverse \(\sin\) button on the calculator (\(\sin^{-1}\)).
\(x = 25.4^\circ\)
Question
The shape ABCDV is a square-based A base with triangular faces meeting at a a point called an apex.. O is the midpoint of the square base ABCD.
Lengths AD, DC, BC and AB are all 4 cm.
The If the angle between two lines is a right angle, the lines are said to be perpendicular. height of the pyramid (OV) is 3 cm.
Calculate the angle between VC and the plane ABCD. Give the answer to 3 significant figures.
ABCD is the base of the pyramid. The line VO and the line OC form a right angle.
Draw the right-angled triangle OVC and label the sides. The angle between VC and the plane is \(y\).
It is not possible to use trigonometry to calculate the angle \(y\) because the length of another side is required.
Pythagoras' theorem can be used to calculate the length OC.
Draw the right-angled triangle ACD and label the sides.
\(c^2 = a^2 + b^2\)
\(AC^2 = 4^2 + 4^2\)
\(AC^2 = 16 + 16\)
\(AC^2 = 32\)
\(AC = \sqrt{32}\)
The length AC is \(\sqrt{32}~cm\)
The point O is in the centre of the length AC so OC is half of the length AC.
The length OC is \(\frac{\sqrt{32}}{2}\) cm.
Use \(\tan{y} = \frac{o}{a}\)
\(\tan{y} = \frac{3}{\frac{\sqrt{32}}{2}}\)
\(\tan{y} = 1.06066 \dotsc\). Do not round this answer yet.
To calculate the angle use the inverse \(\tan\) button on the calculator (\( \tan^{-1}\))
\(y = 46.7^\circ\)
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- Angles, lines and polygons - AQA
- Loci and constructions - AQA
- 2-dimensional shapes - AQA
- 3-dimensional shapes - AQA
- Circles, sectors and arcs - AQA
- Circle theorems - Higher - AQA
- Transformations - AQA
- Pythagoras' theorem - AQA
- Units of measure - AQA
- Vectors - AQA
