IB Syllabus Alignment

Standard Level Topics 3.1–3.3: 3D geometry: distance between points in 3D, midpoint, volume and surface area of right prisms, pyramids, cylinders, cones, and spheres. Right-angled trigonometry: SOH-CAH-TOA, angles of elevation and depression, three-figure bearings. Non-right triangles: the Sine Rule, the Cosine Rule ($c^2 = a^2 + b^2 - 2ab\cos C$ and $\cos C = \frac{a^2+b^2-c^2}{2ab}$), area $A = \frac{1}{2}ab\sin C$, and angles between lines and planes in 3D solids.

1. 3D Coordinates & Volume of Solids (SL 3.1)

Three-dimensional geometry extends 2D Cartesian coordinates by introducing an orthogonal $z$-axis. Distance and midpoint formulas mirror their 2D counterparts with an additional dimension.

  • Distance Between Points in 3D: For points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$
  • Midpoint in 3D: $$M = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}, \; \frac{z_1 + z_2}{2}\right)$$
  • Standard 3D Solids (From IB Formula Booklet):
    • Right Pyramid: $V = \frac{1}{3}Ah$, where $A$ is the base area and $h$ is the perpendicular height.
    • Right Circular Cone: $V = \frac{1}{3}\pi r^2 h$; Curved Surface Area $A = \pi r l$ where slant height $l = \sqrt{r^2 + h^2}$.
    • Sphere: Volume $V = \frac{4}{3}\pi r^3$; Surface Area $A = 4\pi r^2$.
    • Hemisphere: Volume $V = \frac{2}{3}\pi r^3$; Total Surface Area (including flat base) $A = 2\pi r^2 + \pi r^2 = 3\pi r^2$.
Worked Example 1.1 — 3D Distance & Hemisphere Solid

A solid toy consists of a wooden cylinder of radius $r = 4\text{ cm}$ and height $h = 9\text{ cm}$ surmounted by a hemisphere of the same radius.
(a) Find the exact total volume of the solid toy in terms of $\pi$.
(b) Find the total surface area of the toy (including its circular bottom base), giving your answer to 3 significant figures.
(c) Points $A(1, -2, 5)$ and $B(4, 2, -7)$ are marked in a 3D coordinate frame. Find the exact distance $AB$.

  1. Step 1: Calculate total volume: $$V_{\text{cylinder}} = \pi r^2 h = \pi(4^2)(9) = 144\pi\text{ cm}^3$$ $$V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(4^3) = \frac{128\pi}{3}\text{ cm}^3$$ $$V_{\text{total}} = 144\pi + \frac{128\pi}{3} = \frac{432\pi + 128\pi}{3} = \frac{560\pi}{3}\text{ cm}^3 \approx 586\text{ cm}^3$$
  2. Step 2: Calculate total surface area: The total surface consists of: - Bottom circular base: $\pi r^2 = 16\pi$ - Curved cylinder surface: $2\pi r h = 2\pi(4)(9) = 72\pi$ - Curved hemisphere surface: $2\pi r^2 = 2\pi(16) = 32\pi$ $$\text{Total Area} = 16\pi + 72\pi + 32\pi = 120\pi \approx 377\text{ cm}^2$$
  3. Step 3: Calculate 3D distance $AB$: $$d = \sqrt{(4 - 1)^2 + (2 - (-2))^2 + (-7 - 5)^2} = \sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13\text{ units}$$
Answers: (a) $\frac{560\pi}{3}\text{ cm}^3$; (b) $377\text{ cm}^2$ (or $120\pi\text{ cm}^2$); (c) $AB = 13\text{ units}$.

2. Right Triangles, Bearings & Angles of Elevation (SL 3.2)

Real-world navigation and surveying questions rely heavily on right-angled trigonometry, three-figure compass bearings, and vertical angles.

  • Right-Angled Trig Ratios (SOH-CAH-TOA): $$\sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}}, \quad \cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}, \quad \tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}$$
  • Angles of Elevation & Depression:
    • Angle of Elevation: The acute angle measured upward from the horizontal line of sight to an elevated object.
    • Angle of Depression: The acute angle measured downward from the horizontal line of sight to a lower object.
    • Geometric fact: By alternate interior angles, the angle of depression from $A$ to $B$ is identical to the angle of elevation from $B$ to $A$.
  • Three-Figure Compass Bearings:
    • Always measured clockwise from true North ($000^\circ$).
    • Always written with three digits (e.g., $045^\circ$, $090^\circ$, $240^\circ$).
    • Back-bearing rule: If the bearing of $B$ from $A$ is $\theta$, the back-bearing of $A$ from $B$ is $\theta \pm 180^\circ$ ($+180^\circ$ if $\theta < 180^\circ$, $-180^\circ$ if $\theta \ge 180^\circ$).
Worked Example 1.2 — Multi-Point Angle of Elevation

From point $A$ on flat horizontal ground, the angle of elevation to the top of a vertical tower $T$ is $24^\circ$. Walking $50\text{ m}$ directly towards the base of the tower to point $B$, the angle of elevation becomes $38^\circ$.
(a) Set up two trigonometric equations relating the height $h$ of the tower to the horizontal distances.
(b) Calculate the height $h$ of the tower, correct to 3 significant figures.

  1. Step 1: Set up expressions: Let $d$ be the distance from $B$ to the base of the tower. Then the distance from $A$ to the base is $d + 50$. $$\tan 38^\circ = \frac{h}{d} \implies d = \frac{h}{\tan 38^\circ}$$ $$\tan 24^\circ = \frac{h}{d + 50} \implies d + 50 = \frac{h}{\tan 24^\circ}$$
  2. Step 2: Solve simultaneously for $h$: Substitute $d = \frac{h}{\tan 38^\circ}$ into the second equation: $$\frac{h}{\tan 24^\circ} - \frac{h}{\tan 38^\circ} = 50$$ $$h\left(\frac{1}{\tan 24^\circ} - \frac{1}{\tan 38^\circ}\right) = 50$$ Evaluate trigonometric tangents: $\tan 24^\circ \approx 0.44523$, $\tan 38^\circ \approx 0.78129$. $$\frac{1}{0.44523} - \frac{1}{0.78129} \approx 2.2460 - 1.2799 = 0.9661$$ $$h = \frac{50}{0.9661} \approx 51.75 \approx 51.8\text{ m}$$
Answers: (a) $h = d\tan 38^\circ$ and $h = (d + 50)\tan 24^\circ$; (b) $h \approx 51.8\text{ m}$.

3D Spatial Angles & Non-Right Triangles

In 3D solids such as cuboids and pyramids, finding the angle between a spatial line and a plane requires identifying the orthogonal projection of the line onto that plane.

Base Diagonal AC Height h Body Diagonal AG θ A G C Angle Line with Base Plane tan θ = GC / AC 3D Spatial Diagonal AG = √(L² + W² + H²)
Figure 3.3 — 3D Spatial Geometry: Cuboid with body diagonal $AG$, orthogonal base projection $AC$, and spatial angle of elevation $\theta$ satisfying $\tan\theta = \frac{GC}{AC}$.

3. The Cosine & Sine Rules for Non-Right Triangles (SL 3.3)

When solving arbitrary scalene triangles, selecting the correct rule depends on the specific combinations of sides and angles known (SAS, SSS, AAS, or SSA).

  • When to Use the Cosine Rule:
    • SAS (Two sides and the included angle): Finding the third side: $$c^2 = a^2 + b^2 - 2ab\cos C$$
    • SSS (All three sides known): Finding an angle: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ Note: The side being subtracted in the numerator ($c^2$) must be opposite the angle being computed ($C$).
  • When to Use the Sine Rule:
    • AAS / ASA: Finding an unknown side: $$\frac{a}{\sin A} = \frac{b}{\sin B}$$
    • SSA (Known opposite pair + one side): Finding an angle: $$\frac{\sin B}{b} = \frac{\sin A}{a}$$
  • Area of Any Triangle: Given two sides and their included angle: $$\text{Area} = \frac{1}{2}ab\sin C$$
Worked Example 1.3 — SSS Angle & Triangle Area

A triangle has side lengths $a = 7\text{ cm}$, $b = 9\text{ cm}$, and $c = 12\text{ cm}$.
(a) Find the measure of the largest angle in the triangle, giving your answer to the nearest $0.1^\circ$.
(b) Calculate the exact area of the triangle in radical form.

  1. Step 1: Identify the largest angle: The largest angle is always opposite the longest side, which is side $c = 12\text{ cm}$. Thus, angle $C$ is the largest angle.
  2. Step 2: Apply the Cosine Rule for angle $C$: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{7^2 + 9^2 - 12^2}{2(7)(9)} = \frac{49 + 81 - 144}{126} = \frac{130 - 144}{126} = -\frac{14}{126} = -\frac{1}{9}$$ Because $\cos C < 0$, angle $C$ is obtuse! $$C = \arccos\left(-\frac{1}{9}\right) \approx 96.379^\circ \approx 96.4^\circ$$
  3. Step 3: Calculate area: $$\sin C = \sqrt{1 - \cos^2 C} = \sqrt{1 - \left(-\frac{1}{9}\right)^2} = \sqrt{1 - \frac{1}{81}} = \sqrt{\frac{80}{81}} = \frac{\sqrt{80}}{9} = \frac{4\sqrt{5}}{9}$$ $$\text{Area} = \frac{1}{2}ab\sin C = \frac{1}{2}(7)(9)\left(\frac{4\sqrt{5}}{9}\right) = \frac{1}{2}(7)(4\sqrt{5}) = 14\sqrt{5}\text{ cm}^2 \approx 31.3\text{ cm}^2$$
Answers: (a) $C \approx 96.4^\circ$; (b) Exact area $= 14\sqrt{5}\text{ cm}^2$ ($\approx 31.3\text{ cm}^2$).

4. 3D Trigonometry & Bearings Modeling

Worked Example 1.4 — Compass Bearings and Cosine Rule

A lighthouse $L$ is situated $18\text{ km}$ from port $P$ on a bearing of $040^\circ$. A cargo ship $S$ is $25\text{ km}$ from port $P$ on a bearing of $110^\circ$.
(a) Find the angle $\angle LPS$. [1 mark]
(b) Calculate the distance between the ship and the lighthouse to 3 significant figures. [4 marks]
(c) Find the bearing of the ship from the lighthouse. [3 marks]

  1. Part (a): Angle $\angle LPS$: $$\angle LPS = 110^\circ - 040^\circ = 70^\circ \quad [A1]$$
  2. Part (b): Cosine rule on $\triangle LPS$: $$LS^2 = PL^2 + PS^2 - 2(PL)(PS)\cos(\angle LPS) \quad [M1]$$ $$LS^2 = 18^2 + 25^2 - 2(18)(25)\cos(70^\circ) = 324 + 625 - 900(0.34202) \quad [A1]$$ $$LS^2 = 949 - 307.818 = 641.182 \implies LS = \sqrt{641.182} \approx 25.321 \implies 25.3\text{ km} \quad [A1]$$
  3. Part (c): Bearing of ship from lighthouse:

    Find $\angle PLS$ using the sine rule [M1]:

    $$\frac{\sin(\angle PLS)}{25} = \frac{\sin(70^\circ)}{25.321} \implies \sin(\angle PLS) = \frac{25\sin(70^\circ)}{25.321} \approx 0.9278 \implies \angle PLS \approx 68.1^\circ \quad [A1]$$

    Bearing of $P$ from $L$ is $040^\circ + 180^\circ = 220^\circ$. Bearing of $S$ from $L$ is $220^\circ - 68.1^\circ = 151.9^\circ \approx 152^\circ$ [A1].

Answers: (a) $70^\circ$; (b) $25.3\text{ km}$; (c) $152^\circ$.
Worked Example 1.5 — 3D Angle Between a Line and a Plane

A right pyramid has a square base $ABCD$ of side length $8\text{ cm}$. The apex $V$ lies directly above the centre of the base $M$, with vertical height $VM = 10\text{ cm}$.
(a) Find the exact length of the base diagonal $AC$. [2 marks]
(b) Calculate the angle between the slant edge $VA$ and the base plane $ABCD$. [4 marks]

  1. Part (a): Diagonal of square base $ABCD$: $$AC = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}\text{ cm} \quad [M1][A1]$$
  2. Part (b): Angle between edge $VA$ and base:

    The projection of $VA$ onto the base is the segment $[AM]$, where $M$ is the midpoint of diagonal $[AC]$ [M1]:

    $$AM = \frac{1}{2}AC = \frac{8\sqrt{2}}{2} = 4\sqrt{2}\text{ cm} \quad [A1]$$

    In right-angled triangle $\triangle VMA$, the angle $\angle VAM$ satisfies:

    $$\tan(\angle VAM) = \frac{VM}{AM} = \frac{10}{4\sqrt{2}} = \frac{5}{2\sqrt{2}} \approx 1.76777 \quad [M1]$$ $$\angle VAM = \arctan(1.76777) \approx 60.50^\circ \implies 60.5^\circ \quad [A1]$$
Answers: (a) $8\sqrt{2}\text{ cm}$; (b) $60.5^\circ$.

10 Tiered Practice Exercises

Progress from fundamental right-triangle calculations and 3D formulas in Tier 1 to navigation bearings and 3D spatial solids in Tier 2, followed by multi-step optimization and triangulation proofs in Tier 3.

Tier 1: Core Fluency (Problems 1–4)

Exercise 1 Right-Angled Trig & Elevation

A tree casts a shadow of length $18.4\text{ m}$ on horizontal ground when the angle of elevation of the Sun is $52^\circ$. Calculate the height of the tree, correct to 3 significant figures.

View Full Solution

Step 1: Set up trigonometric ratio:
In the right-angled triangle formed by the tree, its shadow, and the sun ray: $$\tan 52^\circ = \frac{\text{Height}}{\text{Shadow length}} = \frac{h}{18.4}$$

Step 2: Solve for $h$:
$$h = 18.4 \times \tan 52^\circ \approx 18.4 \times 1.27994 \approx 23.55 \approx 23.6\text{ m}$$

Final Answer: $23.6\text{ m}$

Exercise 2 3D Distance & Midpoint

Points $P(2, -1, 4)$ and $Q(6, 3, -4)$ are given in 3D Cartesian space.
(a) Find the coordinates of the midpoint of $PQ$.
(b) Find the exact distance between $P$ and $Q$.

View Full Solution

(a) Midpoint $M$:
$$M = \left(\frac{2 + 6}{2}, \; \frac{-1 + 3}{2}, \; \frac{4 + (-4)}{2}\right) = (4, 1, 0)$$

(b) Exact distance:
$$d = \sqrt{(6 - 2)^2 + (3 - (-1))^2 + (-4 - 4)^2} = \sqrt{4^2 + 4^2 + (-8)^2} = \sqrt{16 + 16 + 64} = \sqrt{96} = 4\sqrt{6}\text{ units}$$

Final Answers: (a) $(4, 1, 0)$; (b) $4\sqrt{6}\text{ units}$ ($\approx 9.80$).

Exercise 3 Direct Cosine Rule for Sides

In triangle $ABC$, side $a = 8\text{ cm}$, side $b = 11\text{ cm}$, and angle $C = 42^\circ$. Calculate the length of side $c$, correct to 3 significant figures.

View Full Solution

Step 1: Apply Cosine Rule:
$$c^2 = a^2 + b^2 - 2ab\cos C = 8^2 + 11^2 - 2(8)(11)\cos 42^\circ$$ $$c^2 = 64 + 121 - 176\cos 42^\circ = 185 - 176(0.743145) = 185 - 130.793 = 54.207$$

Step 2: Take square root:
$$c = \sqrt{54.207} \approx 7.3625 \approx 7.36\text{ cm}$$

Final Answer: $7.36\text{ cm}$

Exercise 4 Rearranged Cosine Rule for Angles

In triangle $XYZ$, $x = 5\text{ cm}$, $y = 6\text{ cm}$, and $z = 8\text{ cm}$. Calculate the measure of angle $Z$, correct to the nearest $0.1^\circ$.

View Full Solution

Step 1: Apply rearranged Cosine Rule:
$$\cos Z = \frac{x^2 + y^2 - z^2}{2xy} = \frac{5^2 + 6^2 - 8^2}{2(5)(6)} = \frac{25 + 36 - 64}{60} = \frac{-3}{60} = -0.05$$

Step 2: Solve for $Z$:
Because $\cos Z < 0$, $Z$ is obtuse: $$Z = \arccos(-0.05) \approx 92.866^\circ \approx 92.9^\circ$$

Final Answer: $92.9^\circ$

Tier 2: Exam Modeling (Problems 5–8)

Exercise 5 Triangle Area & Altitude

In triangle $PQR$, $p = 10\text{ cm}$, $q = 14\text{ cm}$, and angle $R = 53^\circ$.
(a) Calculate the area of triangle $PQR$.
(b) Use the cosine rule to calculate side length $r$.
(c) Hence calculate the perpendicular distance from vertex $R$ to side $r$.

View Full Solution

(a) Triangle area:
$$\text{Area} = \frac{1}{2}pq\sin R = \frac{1}{2}(10)(14)\sin 53^\circ = 70\sin 53^\circ \approx 70(0.798636) \approx 55.9\text{ cm}^2$$

(b) Length of side $r$:
$$r^2 = 10^2 + 14^2 - 2(10)(14)\cos 53^\circ = 100 + 196 - 280(0.601815) = 296 - 168.508 = 127.492$$ $$r = \sqrt{127.492} \approx 11.29 \approx 11.3\text{ cm}$$

(c) Perpendicular altitude $h_r$:
$$\text{Area} = \frac{1}{2} \times r \times h_r \implies 55.9045 = \frac{1}{2}(11.291) h_r$$ $$h_r = \frac{2 \times 55.9045}{11.291} \approx 9.90\text{ cm}$$

Final Answers: (a) $55.9\text{ cm}^2$; (b) $11.3\text{ cm}$; (c) $9.90\text{ cm}$.

Exercise 6 Three-Figure Bearings Navigation

A ship sails from port $O$ for $25\text{ km}$ on a bearing of $070^\circ$ to point $A$. It then changes course and sails $18\text{ km}$ on a bearing of $160^\circ$ to point $B$.
(a) Show that angle $O\hat{A}B = 90^\circ$.
(b) Find the direct distance from port $O$ to point $B$.
(c) Calculate the bearing of point $B$ from port $O$.

View Full Solution

(a) Show $\angle OAB = 90^\circ$:
At point $A$, the back-bearing to $O$ is $070^\circ + 180^\circ = 250^\circ$.
The new bearing from $A$ to $B$ is $160^\circ$.
The interior angle inside $\triangle OAB$ at $A$ is $250^\circ - 160^\circ = 90^\circ$.

(b) Distance $OB$:
Since $\angle OAB = 90^\circ$, $\triangle OAB$ is right-angled at $A$: $$OB = \sqrt{OA^2 + AB^2} = \sqrt{25^2 + 18^2} = \sqrt{625 + 324} = \sqrt{949} \approx 30.806 \approx 30.8\text{ km}$$

(c) Bearing of $B$ from $O$:
In $\triangle OAB$: $$\tan(\angle AOB) = \frac{AB}{OA} = \frac{18}{25} = 0.72 \implies \angle AOB = \arctan(0.72) \approx 35.75^\circ$$ Since $OA$ was on bearing $070^\circ$, the bearing to $B$ is: $$070^\circ + 35.75^\circ = 105.75^\circ \approx 106^\circ$$

Final Answers: (a) Interior angle is $250^\circ - 160^\circ = 90^\circ$; (b) $30.8\text{ km}$; (c) Bearing $106^\circ$.

Exercise 7 Right Pyramid Spatial Angles

A right pyramid has a square horizontal base $ABCD$ of side length $12\text{ cm}$. The apex $V$ lies directly above the center $M$ of the base, with vertical height $VM = 15\text{ cm}$.
(a) Calculate the length of the base diagonal $AC$.
(b) Calculate the length of the sloping edge $VA$.
(c) Calculate the angle between the sloping edge $VA$ and the base plane $ABCD$.

View Full Solution

(a) Base diagonal $AC$:
$$AC = \sqrt{12^2 + 12^2} = \sqrt{288} = 12\sqrt{2} \approx 16.97\text{ cm}$$

(b) Length of sloping edge $VA$:
Center $M$ is the midpoint of $AC$, so $AM = \frac{1}{2}AC = 6\sqrt{2}\text{ cm}$.
In right triangle $\triangle VMA$: $$VA = \sqrt{VM^2 + AM^2} = \sqrt{15^2 + (6\sqrt{2})^2} = \sqrt{225 + 72} = \sqrt{297} = 3\sqrt{33} \approx 17.23 \approx 17.2\text{ cm}$$

(c) Angle between $VA$ and base plane:
The angle is $\angle VAM$: $$\tan(\angle VAM) = \frac{VM}{AM} = \frac{15}{6\sqrt{2}} = \frac{5}{2\sqrt{2}} \approx 1.76777$$ $$\angle VAM = \arctan(1.76777) \approx 60.5^\circ$$

Final Answers: (a) $12\sqrt{2}\text{ cm}$ ($\approx 17.0\text{ cm}$); (b) $\sqrt{297}\text{ cm} \approx 17.2\text{ cm}$; (c) $60.5^\circ$.

Exercise 8 Cone Volume & Slant Height

A right circular cone has a base radius of $r = 5\text{ cm}$ and a total volume of $100\pi\text{ cm}^3$.
(a) Find the vertical height $h$ of the cone.
(b) Find the slant height $l$ of the cone.
(c) Find the total surface area of the cone in terms of $\pi$.

View Full Solution

(a) Find vertical height $h$:
$$V = \frac{1}{3}\pi r^2 h \implies 100\pi = \frac{1}{3}\pi(5^2) h \implies 100 = \frac{25}{3}h \implies h = \frac{300}{25} = 12\text{ cm}$$

(b) Find slant height $l$:
$$l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$$

(c) Total surface area:
$$\text{Base Area} = \pi r^2 = \pi(5^2) = 25\pi$$ $$\text{Curved Surface Area} = \pi r l = \pi(5)(13) = 65\pi$$ $$\text{Total Area} = 25\pi + 65\pi = 90\pi\text{ cm}^2 \approx 283\text{ cm}^2$$

Final Answers: (a) $h = 12\text{ cm}$; (b) $l = 13\text{ cm}$; (c) $90\pi\text{ cm}^2$.

Tier 3: Synthesis & Problem-Solving (Problems 9–10)

Exercise 9 Line of Greatest Slope on a Roof Wedge

A triangular right prism $ABC-DEF$ represents a ramp. The horizontal base is rectangle $ABED$ with $AB = 8\text{ m}$ and $BE = 15\text{ m}$. Vertical triangular cross-sections $ABC$ and $DEF$ have $\hat{A} = 90^\circ$ and vertical height $BC = 6\text{ m}$.
(a) Find the length of the sloping edge $AC$ along the triangular cross-section.
(b) Find the length of the diagonal $CD$ on the inclined face $ACFD$.
(c) Calculate the angle between the diagonal $CD$ and the horizontal base plane $ABED$.

View Full Solution

(a) Sloping edge $AC$:
In right triangle $\triangle ABC$: $$AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ m}$$

(b) Diagonal $CD$:
The inclined face $ACFD$ is a rectangle with sides $AC = 10\text{ m}$ and $AD = BE = 15\text{ m}$: $$CD = \sqrt{AC^2 + AD^2} = \sqrt{10^2 + 15^2} = \sqrt{100 + 225} = \sqrt{325} = 5\sqrt{13} \approx 18.03 \approx 18.0\text{ m}$$

(c) Angle between $CD$ and the base plane:
The orthogonal projection of $C$ onto the base plane is point $B$.
The projection of $CD$ on the base plane is line segment $BD$: $$BD = \sqrt{AB^2 + AD^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\text{ m}$$ In right triangle $\triangle CBD$, height $BC = 6\text{ m}$ and base projection $BD = 17\text{ m}$: $$\tan\theta = \frac{BC}{BD} = \frac{6}{17} \approx 0.35294$$ $$\theta = \arctan\left(\frac{6}{17}\right) \approx 19.44^\circ \approx 19.4^\circ$$

Final Answers: (a) $10\text{ m}$; (b) $5\sqrt{13}\text{ m} \approx 18.0\text{ m}$; (c) $19.4^\circ$.

Exercise 10 Multi-Observer Spatial Triangulation

Two observers at ground stations $A$ and $B$, situated $150\text{ m}$ apart on level ground with $B$ on a bearing of $090^\circ$ from $A$, track an aerial drone $D$. From station $A$, the drone is on a horizontal bearing of $040^\circ$ with an angle of elevation of $32^\circ$.
(a) Let $P$ be the point directly beneath the drone on the ground. Find the interior angles of horizontal triangle $\triangle PAB$.
(b) Use the sine rule in $\triangle PAB$ to calculate horizontal distance $AP$.
(c) Calculate the vertical height of the drone above the ground.

View Full Solution

(a) Angles of horizontal triangle $\triangle PAB$:
At station $A$: North is $000^\circ$, $B$ is East at $090^\circ$, drone projection $P$ is at $040^\circ$.
Thus, $\angle PAB = 90^\circ - 40^\circ = 50^\circ$.
If the bearing of $P$ from $B$ is given or observed as $310^\circ$:
At $B$, West (towards $A$) is $270^\circ$. Angle $\angle PBA = 310^\circ - 270^\circ = 40^\circ$.
Third angle: $\angle APB = 180^\circ - (50^\circ + 40^\circ) = 90^\circ$.

(b) Calculate horizontal distance $AP$:
In right triangle $\triangle PAB$ (hypotenuse $AB = 150\text{ m}$): $$\cos(\angle PAB) = \frac{AP}{AB} \implies AP = 150\cos 50^\circ \approx 150(0.642788) \approx 96.418 \approx 96.4\text{ m}$$

(c) Calculate vertical height $h$:
In vertical right triangle $\triangle DAP$: $$\tan 32^\circ = \frac{h}{AP} \implies h = AP\tan 32^\circ \approx 96.418 \times \tan 32^\circ \approx 96.418 \times 0.624869 \approx 60.25 \approx 60.3\text{ m}$$

Final Answers: (a) $\angle PAB = 50^\circ, \angle PBA = 40^\circ, \angle APB = 90^\circ$; (b) $AP \approx 96.4\text{ m}$; (c) Drone height $h \approx 60.3\text{ m}$.

Geometry & Triangle Reference Sheet

In the IB Formula Booklet (Topic 3.1 & 3.3)

The official IB Mathematics AA SL formula booklet lists the following spatial and trigonometric formulas:

Distance in 3D $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
Right Cone Volume & Area $$V = \frac{1}{3}\pi r^2 h, \quad A = \pi r l$$
Sphere Volume & Area $$V = \frac{4}{3}\pi r^3, \quad A = 4\pi r^2$$
Right Pyramid Volume $$V = \frac{1}{3}Ah$$
The Sine Rule $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
The Cosine Rule $$c^2 = a^2 + b^2 - 2ab\cos C$$
Must Memorize & Core Structural Insights
  • Cosine Rule Angle Rearrangement: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$
  • Triangle Area Formula: $$\text{Area} = \frac{1}{2}ab\sin C$$
  • 3D Spatial Line to Plane Angle: $$\tan\theta = \frac{\text{Perpendicular height above plane}}{\text{Length of projection on plane}}$$
  • Back-Bearing Conversion: $$\text{Back-bearing} = \text{Bearing} \pm 180^\circ$$
Critical IB Examination Pitfalls
  • Pitfall 1: Order of Operations in Cosine Rule: Evaluating $a^2 + b^2 - 2ab\cos C$ as $(a^2 + b^2 - 2ab) \times \cos C$. Multiplication precedes subtraction: you must calculate $2ab\cos C$ first, then subtract from $a^2 + b^2$.
  • Pitfall 2: Forgetting Hemisphere Flat Base: When finding the total surface area of a closed hemisphere, omitting the flat circular circular base. Total surface area is $2\pi r^2 + \pi r^2 = 3\pi r^2$.
  • Pitfall 3: Bearings Not Written with 3 Digits: Writing $45^\circ$ instead of $045^\circ$ or measuring counter-clockwise. Bearings must always be 3 digits clockwise from North.
  • Pitfall 4: Misidentifying Angle Between Line and Plane in 3D: Choosing an arbitrary diagonal instead of the orthogonal projection. The projection must drop perpendicular to the plane.

Section A: Diagnostic Multiple Choice

Answer the questions to check your understanding and score.

Score: 0 / 4

Question 1: What is the distance between $P(1, 2, 3)$ and $Q(4, -2, 3)$ in 3D space?

Question 2: In $\triangle ABC$, $a = 4$, $b = 5$, and $C = 60^\circ$. What is the exact length of side $c$?

Question 3: What is the three-figure compass bearing of a direction pointing $30^\circ$ West of North?

Question 4: What is the area of a triangle with sides $6\text{ cm}$ and $10\text{ cm}$ enclosing an angle of $30^\circ$?

Section B: Exam-Style Free-Response Questions

Authentic IB style questions with detailed markschemes showing method ($M$), accuracy ($A$), and reasoning ($R$) marks.

Question 5 (Paper 1 Style) 6 Marks

The diagram shows triangle $ABC$ where $AB = 6\text{ cm}$, $BC = 10\text{ cm}$, and $\cos B = \frac{3}{5}$.
(a) Use the cosine rule to show that $AC = 2\sqrt{13}\text{ cm}$. [2]
(b) Find the exact value of $\sin B$. [1]
(c) Hence find the exact area of triangle $ABC$. [2]
(d) Find the exact value of $\sin A$. [1]

View Full Markscheme

(a) Show $AC = 2\sqrt{13}$: [2 marks]
$$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B \quad (M1)$$ $$AC^2 = 6^2 + 10^2 - 2(6)(10)\left(\frac{3}{5}\right) = 36 + 100 - 120\left(\frac{3}{5}\right) = 136 - 72 = 52$$ $$AC = \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}\text{ cm} \quad (A1)$$

(b) Find $\sin B$: [1 mark]
$$\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \quad (A1)$$

(c) Exact Area of $\triangle ABC$: [2 marks]
$$\text{Area} = \frac{1}{2}(AB)(BC)\sin B = \frac{1}{2}(6)(10)\left(\frac{4}{5}\right) \quad (M1)$$ $$\text{Area} = 30 \times \frac{4}{5} = 24\text{ cm}^2 \quad (A1)$$

(d) Exact $\sin A$: [1 mark]
Using Sine Rule: $$\frac{\sin A}{BC} = \frac{\sin B}{AC} \implies \sin A = \frac{BC \sin B}{AC} = \frac{10(4/5)}{2\sqrt{13}} = \frac{8}{2\sqrt{13}} = \frac{4}{\sqrt{13}} = \frac{4\sqrt{13}}{13} \quad (A1)$$

Question 6 (Paper 2 Style) 8 Marks

A shipping container in the shape of a rectangular cuboid $ABCD-EFGH$ has length $AB = 8\text{ m}$, width $BC = 5\text{ m}$, and height $CG = 3\text{ m}$.
(a) Calculate the length of the base diagonal $AC$. [2]
(b) Calculate the length of the body diagonal $AG$. [2]
(c) Calculate the angle between the body diagonal $AG$ and the horizontal base plane $ABCD$. [2]
(d) A security camera is mounted at vertex $E(0, 0, 3)$ and monitors an item at $M$, the center of face $BCGF$. Find the 3D distance from camera $E$ to item $M$. [2]

View Full Markscheme

(a) Base diagonal $AC$: [2 marks]
$$AC = \sqrt{8^2 + 5^2} = \sqrt{64 + 25} = \sqrt{89} \approx 9.4339 \approx 9.43\text{ m} \quad (M1)(A1)$$

(b) Body diagonal $AG$: [2 marks]
$$AG = \sqrt{8^2 + 5^2 + 3^2} = \sqrt{64 + 25 + 9} = \sqrt{98} \approx 9.899 \approx 9.90\text{ m} \quad (M1)(A1)$$

(c) Angle between $AG$ and base plane $ABCD$: [2 marks]
In right triangle $\triangle ACG$, angle is $\angle GAC$: $$\tan(\angle GAC) = \frac{CG}{AC} = \frac{3}{\sqrt{89}} \approx \frac{3}{9.4339} \approx 0.31798 \quad (M1)$$ $$\angle GAC = \arctan(0.31798) \approx 17.6^\circ \quad (A1)$$

(d) Distance from camera $E$ to $M$: [2 marks]
With $A$ at origin $(0, 0, 0)$: $E = (0, 0, 3)$.
Face $BCGF$ is at $x = 8$, spanning $y \in [0, 5]$ and $z \in [0, 3]$.
Center $M = \left(8, \frac{5}{2}, \frac{3}{2}\right) = (8, 2.5, 1.5)$ (M1).
Distance $EM$: $$d = \sqrt{(8 - 0)^2 + (2.5 - 0)^2 + (1.5 - 3)^2} = \sqrt{64 + 6.25 + 2.25} = \sqrt{72.5} \approx 8.51\text{ m} \quad (A1)$$