Hard GCSE 3D Pythagoras and trigonometry Questions

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 $\tan\theta = \frac{8}{\sqrt{45}}$.

State what the angle between two planes means.

Two planes meet in a line — here the edge $BC$. Pick a point on that line and draw, in each plane, the line through it perpendicular to $BC$. The angle between those two lines is the angle between the planes.

Find the point on BC to work from.

The slant face $VBC$ is isosceles, so the line from $V$ perpendicular to $BC$ meets it at the midpoint $N$. In the base, the line from $M$ perpendicular to $BC$ also meets it at $N$.

Name the triangle.

The two perpendiculars are $NV$ and $NM$, and joining $V$ to $M$ closes the triangle $VMN$, right-angled at $M$ because $VM$ is vertical.

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 $NM$. That is what makes the upright triangle right-angled at $M$.

Rule out the triangle VBC.

This is the slant face itself. It is isosceles, not right-angled, and it contains no part of the base plane to measure against.

Rule out the triangle VMB.

$MB$ runs from the centre to a CORNER, not perpendicular to $BC$, so the angle at $B$ in this triangle is not the angle between the planes.

Rule out the triangle VNC.

This triangle lies inside the slant face and its base $NC$ runs ALONG $BC$. It measures nothing against the base plane.

Rule out the triangle VAM.

Triangle $VAM$ is right-angled, but $AM$ points at a corner. It gives the angle the slant EDGE $VA$ makes with the base, which is smaller than the angle the FACE makes.

Note what the triangle gives you.

Inside triangle $VMN$ it is ordinary 2D trigonometry: $VM = 8$ cm is opposite and $MN = 6$ cm is adjacent.

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.

Note the standard mistake.

Confusing the half-side with the half-diagonal is the commonest error in pyramid questions: the half-side belongs to the FACE angle, the half-diagonal to the EDGE angle.

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 triangle to use.

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

Cut the cone through its axis.

Slicing down through the apex and the centre of the base turns the cone into a flat right-angled triangle: radius across, height up, slant side as the hypotenuse. Every angle in the cone can be read off this triangle.

Mark the angle on the cross-section.

The angle wanted is between the axis of the cone and a slant side. In the cross-section that is the angle shown, and the two sides next to it are the radius and the height — so use tangent.

Substitute the radius and the height.

The opposite side is $r = 8\text{ cm}$ and the adjacent side is $h = 15\text{ cm}$.

Use the inverse tangent to find the angle.

The calculator gives $\theta = 28.0725^\circ$, which rounds to $28.1^\circ$ to $1$ decimal place.

Explain why the cross-section is right-angled.

The axis of the cone is perpendicular to its base, so it meets every radius at a right angle. That is what makes the cross-section a right-angled triangle.

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.

Check the answer against 45 degrees.

The opposite side is shorter than the adjacent side, so the tangent is smaller than $1$ and the angle must be less than $45^\circ$. The answer agrees.

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.

Work out the slant height as a check.

The hypotenuse of the cross-section is the slant height, with $l^2 = 289$; the sine and cosine of the angle could be used instead and would agree.

Note the connected angle.

The angle between the same line and the base is the complement of this one, because the two angles sit in the same right-angled triangle.

Note the standard mistake.

The slant height is the HYPOTENUSE of the cross-section, so it never appears in a tangent. Mixing it in is the commonest error here.

Check the answer against the shape of the cone.

A tall narrow cone gives a steep slant side and a large angle with the base; a wide flat cone gives a small one. The answer matches the numbers.

Check by re-reading the cross-section.

Reading the triangle a second time gives the same two sides in the same order, so the ratio, and the angle, are unchanged.

State the angle.

The angle between the axis of the cone and a slant side is $28.1^\circ$.

State what the angle between two planes means.

The two faces meet along the edge $BC$. Take the point $N$ halfway along $BC$ and draw the line in each face that is perpendicular to $BC$ there: $NV$ in the slant face and $NM$ in the base. The angle between those two lines is the angle between the planes.

Explain why NM and NV are perpendicular to BC.

The face $VBC$ is isosceles ($VB = VC$), so the line from $V$ to the midpoint $N$ is perpendicular to $BC$. In the square base, the line from the centre $M$ to the midpoint of a side is also perpendicular to that side.

Work out the distance MN across the base.

$M$ is the centre of the square and $N$ is the midpoint of $BC$, so $MN$ is half the side: $12$ cm.

Pick the ratio in triangle VMN.

Triangle $VMN$ is right-angled at $M$, with the height $VM = 5$ cm opposite $\theta$ and $MN = 12$ cm adjacent.

Use the inverse tangent to find the angle.

The calculator gives $\theta = 22.6199^\circ$, which rounds to $22.6^\circ$ to $1$ decimal place.

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 $NM$. That is what makes the upright triangle right-angled at $M$.

Label the sides of the right-angled triangle.

Relative to $\theta$, $VM$ is the opposite side and $MN$ is the adjacent side. The hypotenuse is the line itself.

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.

Check the answer against 45 degrees.

The opposite side is shorter than the adjacent side, so the tangent is smaller than $1$ and the angle must be less than $45^\circ$. The answer agrees.

Check with a different ratio.

Using the hypotenuse $VN$ instead, $\sin\theta = \frac{5}{13}$ gives the same angle, so the answer is consistent.

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.

Note the standard mistake.

Using half the DIAGONAL here would give the angle the slant EDGE makes with the base, which is a different (smaller) angle.

Check the slant height.

The hypotenuse of triangle $VMN$ is the slant height, with $VN^2 = 169$.

State the angle.

The angle between the face $VBC$ and the base is $22.6^\circ$.

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 projection of the line on the plane.

Dropping a perpendicular from the far end of $VA$ onto $ABCD$ lands on $M$, so the shadow of $VA$ on the plane is $AM$, and the angle wanted is $\angle AAM$.

Work out the half-diagonal of the base.

The shadow of $VA$ on the base is $AM$, half of the diagonal $AC$ of the square. Halving a length quarters its square, so $AM^2 = 200$.

Draw triangle VMA and pick the ratio.

Triangle $VMA$ is right-angled at $M$. The height $VM = 5$ cm is opposite $\theta$ and the half-diagonal $AM$ is adjacent, so use tangent.

Use the inverse tangent to find the angle.

The calculator gives $\theta = 19.4712^\circ$, which rounds to $19.5^\circ$ to $1$ decimal place.

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 $M$.

Label the sides of the right-angled triangle.

Relative to $\theta$, $VM$ is the opposite side and $AM$ is the adjacent side. The hypotenuse is the line itself.

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.

Keep the intermediate length exact.

Carry $AM^2 = 200$ forward as it is. Rounding $AM$ now and then squaring it again would push a rounding error into the final answer.

Check the answer against 45 degrees.

The opposite side is shorter than the adjacent side, so the tangent is smaller than $1$ and the angle must be less than $45^\circ$. The answer agrees.

Check with a different ratio.

Using the hypotenuse $VA$ instead, $\sin\theta = \frac{1}{3}$ gives the same angle, so the answer is consistent.

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.

Note the standard mistake.

A common error is to use half the SIDE of the square. The edge $VA$ starts at a corner, so the length underneath it is half the DIAGONAL — a bigger number, giving a smaller angle.

State the angle.

The angle between $VA$ and the base is $19.5^\circ$.