GCSE 3D Pythagoras and trigonometry Practice Questions

State the 3D version of Pythagoras theorem.

For a cuboid with edges $a$, $b$ and $c$, the space diagonal $d$ satisfies $d^2 = a^2 + b^2 + c^2$. It comes from using Pythagoras twice: once across the base, then once in the upright triangle.

Substitute the three edges of the cuboid.

The three edges meeting at $A$ are $3$ cm, $4$ cm and $12$ cm. Squaring and adding gives $AG^2 = 169$. Underneath, this is Pythagoras used twice: across the base to get $AC$, then up the triangle $ACG$.

State the length of the diagonal.

The diagonal $AG$ is $13$ cm.

State the 3D version of Pythagoras theorem.

For a cuboid with edges $a$, $b$ and $c$, the space diagonal $d$ satisfies $d^2 = a^2 + b^2 + c^2$. It comes from using Pythagoras twice: once across the base, then once in the upright triangle.

Work out the square of each diagonal.

The diagonal grows with the sum of the squares, so there is no need to take any square roots — just compare the five sums: $166$, $77$, $137$, $102$, $99$.

State the cuboid with the longest diagonal.

The cuboid $6$ cm by $7$ cm by $9$ cm has the longest space diagonal.

State what the angle between a line and a plane means.

The angle between a line and a plane is the angle between the line and its SHADOW on the plane — the projection you get by dropping a perpendicular from the far end of the line straight down onto the plane.

Find where the line meets the plane.

The diagonal $AG$ touches the base at $A$, so the angle must be measured AT $A$. That rules out any angle with its vertex somewhere else.

Find the shadow of the line on the plane.

$G$ is directly above $C$, so the shadow of $AG$ on the base is $AC$. The angle between the line and the plane is therefore the angle between $AG$ and $AC$, which is $\angle GAC$.

Rule out the angle GAB.

$AB$ is an edge of the base, but it is not the shadow of $AG$ — $G$ is not above $B$. $\angle GAB$ is the angle between $AG$ and the edge $AB$, a different angle.

Rule out the angle GAE.

$AE$ is the vertical edge, which is perpendicular to the base rather than in it. $\angle GAE$ is the angle between $AG$ and the VERTICAL, the complement of the angle wanted.

State the angle.

The angle between $AG$ and the base $ABCD$ is $\angle GAC$.

State what the angle between a line and a plane means.

The angle between a line and a plane is the angle between the line and its SHADOW on the plane — the projection you get by dropping a perpendicular from the far end of the line straight down onto the plane.

Drop a perpendicular from the apex onto the base.

The line is $VA$. Its far end is the apex $V$, and $V$ is directly above the centre $M$, so the perpendicular from $V$ lands on $M$. The shadow of $VA$ on the base is $AM$.

Name the triangle made by the line, its shadow and the perpendicular.

The three lines $VA$, $AM$ and $VM$ form triangle $VAM$, right-angled at $M$. That is the triangle to work in.

Explain why the triangle is right-angled at M.

$MV$ is a vertical edge and $ABCD$ is a horizontal plane, so $MV$ is perpendicular to EVERY line drawn in $ABCD$ — including $AM$. That is what makes the upright triangle right-angled at $M$.

Rule out the triangle VAB.

$B$ is a corner of the base, but $V$ is not above $B$, so $AB$ is not the shadow of $VA$. Triangle $VAB$ is the isosceles slant face, not a right-angled triangle at all.

Rule out the triangle VAC.

This triangle does contain the diagonal $AC$, and $M$ lies on it — but the triangle itself is isosceles ($VA = VC$), not right-angled, so no trigonometry can be done in it until $M$ is marked and the triangle is cut in half.

Rule out the triangle VMB.

This triangle does not contain $A$, so it does not contain the line $VA$ and cannot contain the angle wanted.

Rule out the triangle VAD.

$AD$ is an edge of the base, not the shadow of $VA$, and triangle $VAD$ is a slant face, not right-angled.

Check the right angle is at the foot.

$VM$ is vertical and $AM$ lies in the horizontal base, so the angle between them is $90^\circ$. The angle wanted is the one at $A$.

State the triangle to use.

The right-angled triangle to work in is $VAM$.

State what the angle between a line and a plane means.

The angle between a line and a plane is the angle between the line and its SHADOW on the plane — the projection you get by dropping a perpendicular from the far end of the line straight down onto the plane.

Find the shadow of the diagonal on the base.

$G$ is above $C$, so the shadow of $AG$ is $AC$, with $AC^2 = 45$. The angle $\theta$ is $\angle GAC$.

Label the sides of triangle ACG relative to theta.

Triangle $ACG$ is right-angled at $C$. The side opposite $\theta$ is the vertical $CG$, and the side next to it is the base diagonal $AC$.

Choose the right trigonometric ratio.

The two sides that are known are the opposite and the adjacent, so use tangent. Sine or cosine would need the hypotenuse, which has not been worked out yet.

Rule out the upside-down tangent.

This has the opposite and adjacent the wrong way round, so it gives the tangent of the OTHER acute angle in the triangle — the one at $G$.

Rule out the sine statement.

Sine is opposite over HYPOTENUSE, and the hypotenuse of triangle $ACG$ is $AG = \sqrt{109}$ cm, not $8$ cm.

Rule out the cosine statement.

Cosine is adjacent over hypotenuse. The adjacent side is the base DIAGONAL, not the height, so this option has the wrong side on top.

Rule out using an edge instead of the diagonal.

This uses the edge $AB = 3$ cm as the adjacent side. $AB$ is not the shadow of $AG$, so this is the tangent of a different angle.

Explain why the triangle is right-angled at C.

$CG$ is a vertical edge and $ABCD$ is a horizontal plane, so $CG$ is perpendicular to EVERY line drawn in $ABCD$ — including $AC$. That is what makes the upright triangle right-angled at $C$.

Keep the intermediate length exact.

Carry $AC^2 = 45$ forward as it is. Rounding $AC$ now and then squaring it again would push a rounding error into the final answer.

Check the angle is in the right range.

The angle between a line and a plane is always between $0^\circ$ and $90^\circ$, because it is measured inside a right-angled triangle.

Check the calculator is in degrees.

The answer is asked for in degrees, so the calculator must be in degree mode. In radian mode the same inverse tangent would give a completely different number.

Recap the method for the whole topic.

Every question of this kind is solved the same way: find the right-angled triangle hiding inside the solid, redraw it flat, and then do ordinary 2D work in it. Nothing new is needed beyond Pythagoras, sine, cosine and tangent.

Note why a flat triangle is enough.

Any three points lie in a single flat plane, so the triangle can always be redrawn on paper without distortion. That is why a 3D problem collapses to a 2D one as soon as the right three points are chosen.

State the correct statement.

The correct statement is $\tan\theta = \frac{8}{\sqrt{45}}$.