Angles in 3-D Solids

How steep is the sloping edge of a pyramid? What angle does the long diagonal of a box make with the floor? Questions like these come up in architecture and design, and in IB exams. The trick is always the same: find a right triangle hidden inside the solid, flatten it out on paper, and use Pythagoras and trigonometry.

Key ideas

Distance and midpoint in 3-D

For points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, the distance and the midpoint work just like in 2-D, with a third coordinate:

$$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \qquad M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2},\ \frac{z_1 + z_2}{2}\right)$$

Both are in the formula booklet. See vectors in 3-D for more on 3-D coordinates. The distance formula is just Pythagoras used twice; in a cuboid with edges $a$, $b$ and $c$, the long space diagonal has length $\sqrt{a^2 + b^2 + c^2}$.

The angle between two lines that meet

Two lines that intersect lie in one flat plane, so the angle between them is an ordinary angle in a triangle. Find a triangle that contains both lines, work out the side lengths you need, and use trigonometry. If the triangle has a right angle, SOH CAH TOA is enough.

The angle between a line and a plane

Suppose a line meets a plane at a point $A$. Pick another point $P$ on the line and drop a perpendicular from $P$ straight down to the plane, meeting it at $F$ (the foot of the perpendicular).

• The line $AF$ is the projection of the line onto the plane: its “shadow” when light shines straight down.
• The angle between the line and the plane is the angle $\angle PAF$ between the line and its projection.
• Triangle $PAF$ always has a right angle at $F$, so

$$\tan \theta = \frac{PF}{AF}, \qquad \sin \theta = \frac{PF}{AP}, \qquad \cos \theta = \frac{AF}{AP}$$

In a cuboid the foot is usually a corner (the point directly below), and in a right pyramid or cone it’s the centre of the base.

In each solid, the line, its projection on the base, and a vertical line make a right triangle.

A step-by-step plan

1. Sketch the solid and label it. Mark the line (and plane) you care about.
2. Identify the right triangle: the line, its projection, and the vertical (perpendicular) side.
3. Find any missing side lengths, often with Pythagoras in a different right triangle (like the base diagonal $AC$).
4. Redraw the right triangle flat, then use a trig ratio.

In SL exams, 3-D questions only need right-angled trigonometry. Keep full calculator values and give angles in degrees to 3 s.f. Later you’ll meet a vector method for the same angles in angles between lines and planes.

Worked examples

Example 1: The diagonal of a cuboid and the base

The cuboid $ABCDEFGH$ in the figure has $AB = 8$ cm, $BC = 6$ cm and $AE = 5$ cm. The base is $ABCD$, and $E$, $F$, $G$, $H$ are directly above $A$, $B$, $C$, $D$. Find the length of $AG$ and the angle between $AG$ and the base.

Solution. $G$ is directly above $C$, so the projection of $AG$ on the base is $AC$, and triangle $ACG$ has a right angle at $C$.

First find $AC$ in the right triangle $ABC$ on the base:

$$AC = \sqrt{8^2 + 6^2} = 10 \text{ cm}$$

Then in triangle $ACG$, with $CG = 5$:

$$AG = \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.2 \text{ cm} \qquad \tan \theta = \frac{CG}{AC} = \frac{5}{10} \quad\Rightarrow\quad \theta = 26.565\ldots^\circ \approx 26.6^\circ$$

Check: $\sqrt{8^2 + 6^2 + 5^2} = \sqrt{125}$, matching the space-diagonal formula.

Example 2: The angle between two lines

In the same cuboid, find the angle between the lines $AG$ and $AB$.

Solution. Both lines start at $A$, so look at triangle $ABG$. The edge $AB$ is perpendicular to the whole face $BCGF$, so it’s perpendicular to $BG$, which lies in that face. That puts the right angle at $B$.

We know the side next to the angle ($AB = 8$) and the hypotenuse ($AG = \sqrt{125}$):

$$\cos(\angle GAB) = \frac{AB}{AG} = \frac{8}{\sqrt{125}} \quad\Rightarrow\quad \angle GAB = 44.312\ldots^\circ \approx 44.3^\circ$$

Example 3: The edge of a pyramid

The right pyramid $VABCD$ in the figure has a square base of side $10$ m, and $V$ is $12$ m directly above the centre $O$ of the base. Find

• (a) the angle between the edge $VA$ and the base,
• (b) the length of $VA$,
• (c) the angle between the base and the line $VM$, where $M$ is the midpoint of $BC$.

Solution. (a) The projection of $VA$ on the base is $OA$, which is half of the diagonal $AC$:

$$AC = \sqrt{10^2 + 10^2} = 10\sqrt{2} \qquad OA = 5\sqrt{2}$$

Triangle $VOA$ has a right angle at $O$:

$$\tan \alpha = \frac{VO}{OA} = \frac{12}{5\sqrt{2}} \quad\Rightarrow\quad \alpha = 59.491\ldots^\circ \approx 59.5^\circ$$

(b)

$$VA = \sqrt{12^2 + (5\sqrt{2})^2} = \sqrt{144 + 50} = \sqrt{194} \approx 13.9 \text{ m}$$

(c) The projection of $VM$ is $OM$, and $OM = 5$ (half the side length). In the right triangle $VOM$:

$$\tan \beta = \frac{12}{5} \quad\Rightarrow\quad \beta = 67.380\ldots^\circ \approx 67.4^\circ$$

Notice that $VM$ is steeper than the edge $VA$: it reaches the base sooner.

Example 4: Using coordinates

The ground is the $xy$-plane, with units in metres. A cable runs in a straight line from an anchor $A(2, 1, 0)$ on the ground to the top of a mast at $T(5, 5, 6)$. Find the length of the cable, the coordinates of its midpoint, and the angle the cable makes with the ground.

Solution. Length and midpoint:

$$AT = \sqrt{(5 - 2)^2 + (5 - 1)^2 + (6 - 0)^2} = \sqrt{9 + 16 + 36} = \sqrt{61} \approx 7.81 \text{ m}$$ $$M = \left(\frac{2 + 5}{2},\ \frac{1 + 5}{2},\ \frac{0 + 6}{2}\right) = (3.5,\ 3,\ 3)$$

The point on the ground directly below $T$ is $F(5, 5, 0)$, so the projection of $AT$ is $AF$:

$$AF = \sqrt{3^2 + 4^2} = 5 \qquad TF = 6 \qquad \tan \theta = \frac{6}{5} \quad\Rightarrow\quad \theta = 50.194\ldots^\circ \approx 50.2^\circ$$

Common mistakes

Using the wrong side of the base as the projection. In Example 1, the projection of $AG$ is the diagonal $AC$, not the edge $AB$ or $AD$. Always ask “which point is directly below the top end of my line?” and join it to the bottom end.

Putting the right angle in the wrong place. The right angle is at the foot of the perpendicular ($C$ in Example 1, $O$ in Example 3), never at the point where the line meets the plane. If your “right triangle” has the right angle at the angle you want, start again.

Using an edge length where a half-diagonal is needed. For the edge of a square-based pyramid, the horizontal distance is half the base diagonal ($5\sqrt{2}$ in Example 3), not half the side ($5$). Half the side is the right distance for the slant height $VM$.

Finding the wrong angle in the triangle. Once the triangle is flat, label the angle you want and decide which sides are opposite and adjacent to that angle. In Example 1, $\tan\theta = \dfrac{5}{10}$, not $\dfrac{10}{5}$.

Rounding the intermediate length. If you round $AC$ or $OA$ and then use it again, your final angle can be off in the third significant figure. Store exact values like $5\sqrt{2}$ or full calculator values.

Forgetting the third coordinate. In 3-D the distance formula has three squared differences. Leaving out the $z$-difference gives the length of the shadow, not of the line.

Practice

1. (Warm-up) Find the distance between $P(1, -2, 3)$ and $Q(5, 2, -1)$, and the midpoint of $PQ$.

Solution

$$PQ = \sqrt{4^2 + 4^2 + (-4)^2} = \sqrt{48} = 4\sqrt{3} \approx 6.93$$$$M = \left(\frac{1 + 5}{2},\ \frac{-2 + 2}{2},\ \frac{3 + (-1)}{2}\right) = (3,\ 0,\ 1)$$

2. (Warm-up) A right cone has base radius $4$ cm and height $9$ cm. Find the angle between a slant edge of the cone and its base.

Solution

The height, the radius and the slant edge form a right triangle with the right angle at the centre of the base. The radius is the projection of the slant edge:

$$\tan \theta = \frac{9}{4} \quad\Rightarrow\quad \theta = 66.037\ldots^\circ \approx 66.0^\circ$$

3. (Warm-up) A cube has edges of length $4$ cm. Find the angle between a space diagonal of the cube and its base.

Solution

The projection of the space diagonal is a diagonal of the base: $\sqrt{4^2 + 4^2} = 4\sqrt{2}$. The vertical side is an edge, $4$:

$$\tan \theta = \frac{4}{4\sqrt{2}} = \frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta = 35.264\ldots^\circ \approx 35.3^\circ$$

(Every cube gives the same angle, whatever its size.)

4. (Core) The cuboid $ABCDEFGH$ has base $ABCD$ with $AB = 12$ cm and $BC = 5$ cm, and height $AE = 4$ cm. $E$, $F$, $G$, $H$ are directly above $A$, $B$, $C$, $D$.

• (a) Find the length of $AG$.
• (b) Find the angle between $AG$ and the base $ABCD$.
• (c) Find the angle between $AG$ and the face $ABFE$.

Solution

(a)

$$AG = \sqrt{12^2 + 5^2 + 4^2} = \sqrt{185} \approx 13.6 \text{ cm}$$

(b) The projection on the base is $AC = \sqrt{12^2 + 5^2} = 13$, and $CG = 4$:

$$\tan \theta = \frac{4}{13} \quad\Rightarrow\quad \theta = 17.102\ldots^\circ \approx 17.1^\circ$$

(c) The edge $GF$ is perpendicular to the face $ABFE$, so $F$ is the foot of the perpendicular from $G$, and the projection of $AG$ on that face is $AF$. Here $GF = BC = 5$ and

$$AF = \sqrt{12^2 + 4^2} = \sqrt{160}$$$$\tan \phi = \frac{5}{\sqrt{160}} \quad\Rightarrow\quad \phi = 21.568\ldots^\circ \approx 21.6^\circ$$

5. (Core) A right pyramid $VABCD$ has a rectangular base with $AB = 8$ m and $BC = 6$ m. The apex $V$ is $10$ m above the centre $O$ of the base.

• (a) Find the length of the edge $VA$.
• (b) Find the angle between $VA$ and the base.
• (c) Find the angle $\angle AVC$.

Solution

(a) The base diagonal is $AC = \sqrt{8^2 + 6^2} = 10$, so $OA = 5$.

$$VA = \sqrt{10^2 + 5^2} = \sqrt{125} \approx 11.2 \text{ m}$$

(b) In the right triangle $VOA$:

$$\tan \theta = \frac{10}{5} = 2 \quad\Rightarrow\quad \theta = 63.434\ldots^\circ \approx 63.4^\circ$$

(c) Triangle $AVC$ is isosceles ($VA = VC$), and $VO$ splits it into two right triangles. In triangle $VOA$, the angle at $V$ satisfies

$$\tan(\angle AVO) = \frac{OA}{VO} = \frac{5}{10} \quad\Rightarrow\quad \angle AVO = 26.565\ldots^\circ$$$$\angle AVC = 2 \times 26.565\ldots^\circ = 53.130\ldots^\circ \approx 53.1^\circ$$

6. (Core) A flagpole stands on flat ground, which is the $xy$-plane (units in metres). Its base is at $F(6, 8, 0)$ and its top is at $T(6, 8, 12)$. A rope runs straight from $T$ to a peg at $P(1, -4, 0)$.

• (a) Find the length of the rope.
• (b) Find the coordinates of the midpoint of the rope.
• (c) Find the angle between the rope and the ground.

Solution

(a)

$$PT = \sqrt{(6 - 1)^2 + (8 - (-4))^2 + 12^2} = \sqrt{25 + 144 + 144} = \sqrt{313} \approx 17.7 \text{ m}$$

(b)

$$M = \left(\frac{1 + 6}{2},\ \frac{-4 + 8}{2},\ \frac{0 + 12}{2}\right) = (3.5,\ 2,\ 6)$$

(c) The projection of the rope on the ground is $PF$:

$$PF = \sqrt{5^2 + 12^2} = 13 \qquad \tan \theta = \frac{12}{13} \quad\Rightarrow\quad \theta = 42.709\ldots^\circ \approx 42.7^\circ$$

7. (Core) In the cuboid from Example 1 ($AB = 8$, $BC = 6$, $AE = 5$), find the angle between the face diagonal $AF$ and the base $ABCD$.

Solution

$F$ is directly above $B$, so the projection of $AF$ on the base is $AB = 8$, and $BF = 5$. Triangle $ABF$ has its right angle at $B$:

$$\tan \theta = \frac{5}{8} \quad\Rightarrow\quad \theta = 32.005\ldots^\circ \approx 32.0^\circ$$

8. (Challenge) A right pyramid has a square base of side $6$ m. Each sloping edge makes an angle of $60^\circ$ with the base. Find the exact height of the pyramid and its volume.

Solution

Half the base diagonal is $\dfrac{6\sqrt{2}}{2} = 3\sqrt{2}$. In the right triangle formed by the height, the half-diagonal and an edge:

$$\tan 60^\circ = \frac{h}{3\sqrt{2}} \quad\Rightarrow\quad h = 3\sqrt{2} \times \sqrt{3} = 3\sqrt{6} \approx 7.35 \text{ m}$$$$V = \frac{1}{3}(6^2)(3\sqrt{6}) = 36\sqrt{6} \approx 88.2 \text{ m}^3 \quad (\text{3 s.f.})$$

9. (Challenge) A right cone has slant height $10$ cm. The angle between two slant edges on opposite sides of the cone (through the apex, in a vertical cross-section) is $50^\circ$. Find the radius, the height and the volume of the cone.

Solution

The vertical cross-section is an isosceles triangle with two sides $10$ and apex angle $50^\circ$. The height splits it into two right triangles, each with an angle of $25^\circ$ at the apex and hypotenuse $10$:

$$r = 10 \sin 25^\circ = 4.2261\ldots \approx 4.23 \text{ cm} \qquad h = 10 \cos 25^\circ = 9.0630\ldots \approx 9.06 \text{ cm}$$$$V = \frac{1}{3}\pi r^2 h = 169.51\ldots \approx 170 \text{ cm}^3 \quad (\text{3 s.f.})$$