Trigonometry in three dimensions: finding the right-angled triangle inside a box or pyramid, space diagonals, the angle between a line and a plane, angles of elevation with bearings, and the sine and cosine rules on a horizontal plane (moved into Extension 1 in the 2024 syllabus)

Work in two right triangles. First find the base diagonal $AC$ in triangle $ABC$ (right-angled at $B$):

Then $CG$ is vertical, so triangle $ACG$ is right-angled at $C$ with $CG = AE = 5$:

Marker's note: one mark for $AC$ and one for $AG$. The shortcut $AG = \sqrt{6^2 + 8^2 + 5^2}$ is also accepted.

The projection of $AG$ onto the base is $AC$, because $G$ is directly above $C$. So the required angle is $\angle GAC$ in the right-angled triangle $ACG$.

Use tangent, with opposite side $CG = 5$ and adjacent side $AC = 10$:

Marker's note: one mark for naming the projection $AC$, one for the correct ratio, one for the angle. Using $AG$ as the adjacent side in a tangent ratio is the typical slip.

The half-diagonal of the base. The diagonal of the square is $8\sqrt{2}$, so $AM = 4\sqrt{2}$.

Slant edge. Triangle $VMA$ is right-angled at $M$:

Angle with the base. The projection of $VA$ onto the base is $AM$, so

Marker's note: one mark for $AM$, one for $VA$, one for the angle. The half-diagonal, not the half-side, is the horizontal distance from $A$ to the centre.

Use the midpoint of $AB$. Let $N$ be the midpoint of $AB$. Then $VN \perp AB$ (the face is isosceles) and $MN \perp AB$ (the base is a square), so $\angle VNM$ is the angle between the face and the base.

The horizontal distance. $MN$ is half a side of the square: $MN = 4$ m.

Use tangent in the right-angled triangle $VMN$:

Marker's note: one mark for identifying the correct right-angled triangle (the line $VN$ perpendicular to the edge), one for the ratio, one for the angle. The slant height $VN = \sqrt{36 + 16} = \sqrt{52}$ is not needed here.

Let the height be $h$ and the base of the tower be $O$. In the vertical right triangles,

Triangle $AOB$ is right-angled at $O$ because $A$ is due south and $B$ is due east of $O$. By Pythagoras,

Solve for $h$.

As a check, $OA \approx 87.4$ m and $OB \approx 48.6$ m, and $87.4^2 + 48.6^2 \approx 10\,000$.

Marker's note: one mark for each of the two expressions for $OA$ and $OB$, one for the right angle at $O$ and Pythagoras, one for $h$.

Find the two ground distances from the vertical right triangles.

Apply the cosine rule in the horizontal triangle $ATB$.

Substituting, $AB^2 \approx 9802 + 4432 + 4508 \approx 18\,742$, so $AB \approx 137$ m.

Marker's note: one mark for each of $TA$ and $TB$, one for the cosine rule setup (note $\cos 110^\circ$ is negative so the third term adds), one for the distance. Keep the full calculator values until the end to avoid rounding drift.

Find the projection of $AG$ onto the plane $ABFE$. The edge $GF$ is parallel to $BC$, so it is perpendicular to the face $ABFE$. Hence the foot of the perpendicular from $G$ to the plane is $F$, and the projection of $AG$ onto the plane is $AF$. The required angle is $\angle GAF$.

Lengths in triangle $AFG$ (right-angled at $F$).

$FG = BC = 8$ and $AF = \sqrt{AB^2 + BF^2} = \sqrt{36 + 25} = \sqrt{61}$.

Angle.

Check with sine. $\sin\angle GAF = \frac{FG}{AG} = \frac{8}{\sqrt{125}}$ gives the same angle.

Marker's note: one mark for identifying the projection $AF$, one for $FG$ and $AF$, one for the ratio, one for the angle. Using the base diagonal $AC$ here would find the angle with the wrong plane.

(a) Edge and base. The diagonal of the base is $\sqrt{12^2 + 8^2} = \sqrt{208}$, so $AM = \frac{\sqrt{208}}{2} \approx 7.21$ m. In the right triangle $VMA$,

(b) Face and base. Let $N$ be the midpoint of $BC$. Then $VN \perp BC$ and $MN \perp BC$, and $BC$ is the side of length $8$. The distance from the centre to this side is $MN = \frac{AB}{2} = 6$ m, so

Marker's note: two marks for each part. In (b) the distance from the centre to the side $BC$ is half the other side length, $AB \div 2$, which is the usual place to slip.