What do you look for first in any 3D Pythagoras or trigonometry problem?

A right-angled triangle inside the solid, with two of its three sides already known. Redrawing that triangle flat, away from the 3D picture, makes it far easier to work with.

How can a 3D Pythagoras problem be solved without using the 3D formula?

Break it into two 2D problems by finding two right-angled triangles, and apply ordinary Pythagoras to each in turn. The first triangle gives a diagonal across one face, and that diagonal then becomes a side of the second triangle.

IGCSEMathsEdexcelMaths BFlashcardsTrigonometry3D Pythagoras & Trigonometry

3D Pythagoras & Trigonometry (Edexcel IGCSE Maths B): Flashcards

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What do you look for first in any 3D Pythagoras or trigonometry problem?

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What do you look for first in any 3D Pythagoras or trigonometry problem?
A right-angled triangle inside the solid, with two of its three sides already known.
Redrawing that triangle flat, away from the 3D picture, makes it far easier to work with.
Complete the 3D version of Pythagoras' theorem.
$d^{2} = x^{2} +______+ z^{2}$
The completed formula is:
$d^{2} = x^{2} + y^{2} + z^{2}$
Here $d$ is the distance between two points, and $x$ , $y$ and $z$ are the distances between them in the three perpendicular directions.
How can a 3D Pythagoras problem be solved without using the 3D formula?
Break it into two 2D problems by finding two right-angled triangles, and apply ordinary Pythagoras to each in turn.
The first triangle gives a diagonal across one face, and that diagonal then becomes a side of the second triangle.
A box measures 3 cm by 4 cm by 6 cm. How long is the longest pencil that will fit inside it?
The longest pencil fits between diagonally opposite corners, so $d^{2} = 3^{2} + 4^{2} + 6^{2} = 61$ .
That gives $d = \sqrt{61} = 7.81$ cm to 3 significant figures.
True or False?
A cone's slant height can be found from its radius and its perpendicular height using Pythagoras.
True.
The radius, the perpendicular height and the slant height form a right-angled triangle, with the slant height as the hypotenuse.
So the slant height $l$ satisfies $l^{2} = r^{2} + h^{2}$ .
How do you find the angle between a cone's slant height and its base?
Use the right-angled triangle formed by the radius, the perpendicular height and the slant height, then apply SOHCAHTOA to it.
The radius is adjacent to that angle and the height is opposite it, so $\tan$ of the angle is the height divided by the radius.

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