HG20 - Three-dimensional trigonometry | Eduqas GCSE Mathematics

Turn a problem in a solid into a flat triangle with justified lengths and right angles. Use successive Pythagoras or trigonometry calculations, then identify the exact angle requested. This is the Higher three-dimensional contribution to HG20.

Finding right triangles

A cuboid measures $8\,\mathrm{cm}$ by $6\,\mathrm{cm}$ by $4\,\mathrm{cm}$ . How long is a straight internal diagonal joining opposite corners? The answer needs a triangle passing through the interior, not just a rectangular face.

A plane is a flat surface extending in all directions. A triangle's three vertices lie in one plane, even when the triangle cuts through a solid. A line perpendicular to a plane meets every line in that plane through its foot at a right angle.

To project a point onto a plane, drop the perpendicular to the plane. Projecting the ends of a segment gives its perpendicular projection onto that plane. In a cuboid, vertical edges are perpendicular to its horizontal base.

[DIAGRAM: asset_name: A diagonal and its base projection; asset_slug: cuboid-diagonal-section; caption: The internal diagonal AG, base projection AC and vertical edge CG form a right triangle.; description: Perspective cuboid ABCDEFGH with A B C D around the bottom and E F G H directly above, dimensions AB 8 cm BC 6 cm CG 4 cm. The ACG section is highlighted beside its true-scale flat right triangle, right-angled at C.]
The internal diagonal AG, base projection AC and vertical edge CG form a right triangle.

Use a perspective sketch to identify edges and planes, then redraw the selected triangle flat. Apparent angles on a perspective drawing are not reliable measurements. State why the chosen angle is $90^\circ$ .

Worked example 1: a cuboid projection

Question. In the cuboid shown, $AB=8\,\mathrm{cm}$ , $BC=6\,\mathrm{cm}$ , $CG=4\,\mathrm{cm}$ . Identify the base projection of $AG$ , calculate its length, and identify a right triangle containing $AG$ .

Plan. Project $G$ vertically to $C$ ; $A$ already lies in the base.

AC^2=8^2+6^2=100,\qquad AC=10\,\mathrm{cm}.

Answer. The projection is $AC=10\,\mathrm{cm}$ . Triangle $ACG$ contains $AG$ and is right-angled at $C$ , since $CG$ is perpendicular to the base and $AC$ lies in that base.

Check. The other base diagonal $BD$ is also $10\,\mathrm{cm}$ , as opposite sides of the base rectangle have the same lengths. A base diagonal is longer than either $8\,\mathrm{cm}$ or $6\,\mathrm{cm}$ .

In a right square-based pyramid, the apex lies directly above the base centre. An edge and a face's sloping height have different base projections.

[DIAGRAM: asset_name: Pyramid base projections and sections; asset_slug: pyramid-right-sections; caption: An edge projects to a half-diagonal; the sloping face height projects to a centre-to-midpoint segment.; description: Plan of a 10 cm square base ABCD with centre O and M midpoint of AB, showing AO 5 square root 2 cm and OM 5 cm. Beside it are true-scale right sections SOA and SOM with SO 12 cm, right angles at O, and their different horizontal lengths.]
An edge projects to a half-diagonal

Worked example 2: two pyramid sections

Question. A right square-based pyramid has base side $10\,\mathrm{cm}$ . Its apex $S$ is $12\,\mathrm{cm}$ directly above centre $O$ . Let $M$ be the midpoint of base edge $AB$ . Identify the projections of $SA$ and $SM$ , and find the projected lengths.

Plan. Project $S$ to $O$ . Locate $A$ and $M$ in the square base.

The whole base diagonal is $\sqrt{10^2+10^2}=10\sqrt2\,\mathrm{cm}$ . Its centre bisects it, while the distance from the centre to an edge midpoint is half a side.

AO=5\sqrt2\,\mathrm{cm},\qquad OM=5\,\mathrm{cm}.

Answer. $SA$ projects to $AO$ ; $SM$ projects to $OM$ . Triangles $SOA$ and $SOM$ are both right-angled at $O$ .

Check. In the base, $AM=5\,\mathrm{cm}$ and $OM=5\,\mathrm{cm}$ are perpendicular, so $AO^2=5^2+5^2=50$ , confirming $AO=5\sqrt2\,\mathrm{cm}$ .

The projection depends on the actual endpoints of the line, not on which base edge looks nearest on the page.

Watch out

Half a square's diagonal is not half its side. For an apex-to-corner edge use the centre-to-corner distance; for an apex-to-edge-midpoint line use the centre-to-midpoint distance.

Choose the two projections in a new pyramid.

Lengths in 3D

Solve the base or face triangle first if it supplies a side of the interior triangle. Carry the exact square root or unrounded calculator value into the next stage.

The current Higher additional formula sheet supplies Pythagoras and the right-triangle sine, cosine and tangent ratios. Choosing a valid triangle and its side roles still requires geometric reasoning.

Worked example 3: the internal diagonal

Question. Find the internal diagonal $AG$ of the $8\,\mathrm{cm}$ by $6\,\mathrm{cm}$ by $4\,\mathrm{cm}$ cuboid to $3$ significant figures.

Plan. First find base diagonal $AC$ , then use right triangle $ACG$ .

AC=\sqrt{8^2+6^2}=10\,\mathrm{cm}, \qquad AG=\sqrt{10^2+4^2}=\sqrt{116}=2\sqrt{29}\,\mathrm{cm}.

Answer. $AG=10.8\,\mathrm{cm}$ to $3$ significant figures.

Check. Take a different route. Face diagonal $BG=\sqrt{6^2+4^2}=\sqrt{52}\,\mathrm{cm}$ . Since $AB$ is perpendicular to face $BCGF$ , triangle $ABG$ is right-angled at $B$ ; it gives $AG^2=8^2+52=116$ again.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Number

Algebra

Ratio and proportion

Geometry and measures

Higher graphs

Probability

Statistics

HG20 - Three-dimensional trigonometry

Finding right triangles

Lengths in 3D

Continue this lesson.

Learn faster with Nova