AQA · Mathematics · 8300
GCSE Maths: Trigonometry: 3D
Geometry & Measures · AQA GCSE Maths (8300). Cards are generated from the Corbettmaths practice questions for Trigonometry: 3D (Corbettmaths Video 332), and every card links back to the exact page it came from. Practice tests generate fresh variations of these questions, so the numbers change every attempt. Source material © Corbettmaths — used with attribution.
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In 3D trigonometry problems involving pyramids, where is the apex typically positioned relative to the base?
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Directly over the centre of the base.
In a square-based pyramid where the side of the square is s, the diagonal across the base has length ?.
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s√2
In a square-based pyramid, what is the first step to find the perpendicular height when given the base side length and slant edge length?
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Calculate the diagonal of the square base, then use it with the slant edge in a right triangle (Pythagoras) where the height is perpendicular.
What formula calculates the volume of a pyramid?
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V = (1/3) × base area × height
In a rectangular-based pyramid with the apex over the centre, how do you find the distance from the centre to a corner of the base?
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Use Pythagoras with half the length and half the width: distance = √((length/2)² + (width/2)²)
To find the angle between a triangular face and the base of a pyramid, what key element do you identify?
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The perpendicular from the apex to an edge of the base. This creates a right triangle where you can use trigonometry to find the angle.
In a 3D problem involving a cuboid, how do you find the diagonal across the base rectangle?
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Use Pythagoras' theorem: diagonal = √(length² + width²)
When finding an angle in a 3D problem, why is it important to identify a right-angled triangle?
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Because trigonometric ratios (sin, cos, tan) only apply directly in right-angled triangles, allowing you to calculate angles from side lengths.
In a pyramid problem where M is the midpoint of an edge, how does this simplify distance calculations?
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It allows you to work with half the edge length, creating smaller right triangles that are easier to solve using Pythagoras or trigonometry.
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