MYP Inquiry Framework
Key Concept Relationships
Related Concepts Measurement, Perspective, Models
Global Context Orientation in space and time (Geodetic surveying & astronomical navigation)
Statement of Inquiry Invariant trigonometric ratios allow observers to measure inaccessible spatial distances through angle relationships.
2.1

The Primary Trigonometric Ratios (SOH CAH TOA)

  • Definitions in a Right-Angled Triangle: With reference to an acute angle $\theta$: $$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \quad (\text{SOH})$$ $$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \quad (\text{CAH})$$ $$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \quad (\text{TOA})$$
  • Quotient & Fundamental Pythagorean Identity: $$\tan \theta = \frac{\sin \theta}{\cos \theta}, \qquad \sin^2 \theta + \cos^2 \theta = 1$$
  • Inverse Trigonometric Functions: To find an unknown angle when two sides are known, use the inverse operations $\sin^{-1}$, $\cos^{-1}$, or $\tan^{-1}$: $$\theta = \tan^{-1}\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)$$
45° - 45° - 90° Triangle 1 1 √2 45° 30° - 60° - 90° Triangle √3 1 2 30° 60°
2.2

Exact Values for Special Angles ($30^\circ, 45^\circ, 60^\circ$)

  • Summary Table of Exact Ratios:
    $\theta$ $\sin \theta$ $\cos \theta$ $\tan \theta$
    $30^\circ$ $\frac{1}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
    $45^\circ$ $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $1$
    $60^\circ$ $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $\sqrt{3}$
  • Complementary Angle Relationships: $$\sin(90^\circ - \theta) = \cos \theta, \qquad \cos(90^\circ - \theta) = \sin \theta$$
2.3

Angles of Elevation, Depression & Bearings

  • Angle of Elevation: The angle measured upwards from the horizontal eye-level to the object above.
  • Angle of Depression: The angle measured downwards from the horizontal eye-level to the object below. Geometric fact: The angle of depression from $A$ to $B$ equals the angle of elevation from $B$ to $A$ (alternate interior angles across parallel horizontals).
  • Three-Figure Bearings: A direction given as an angle measured clockwise from True North, written with three digits (e.g., $045^\circ, 130^\circ, 285^\circ$). $$\text{Back Bearing} = \text{Bearing} \pm 180^\circ$$
2.4

3D Trigonometry & Angles between Lines and Planes

  • Angle Between a Line and a Plane: The angle $\theta$ between a line $AB$ and a plane is the angle between $AB$ and its orthogonal projection $AB'$ onto that plane.
    1. Drop a perpendicular from vertex $B$ to the plane, hitting at $B'$.
    2. Connect $A$ and $B'$ along the plane.
    3. Solve the right-angled triangle $\triangle AB B'$ where $\angle AB'B = 90^\circ$: $$\sin \theta = \frac{BB'}{AB}, \qquad \tan \theta = \frac{BB'}{AB'}$$

Worked Examples (Criterion A & C)

Example 1 Two-Point Inaccessible Height Problem

Problem: A surveyor measures the angle of elevation to the top of a communications tower as $28^\circ$. Walking $50\text{ metres}$ directly closer to the base on level ground, the angle of elevation increases to $42^\circ$. Find the height of the tower to 3 significant figures.

Step 1: Set up simultaneous trigonometric equations

Let $h$ be tower height, and $x$ be distance from second observation point to base.

From the second position: $\tan(42^\circ) = \frac{h}{x} \implies h = x \tan(42^\circ)$.

From the first position: $\tan(28^\circ) = \frac{h}{x + 50} \implies h = (x + 50)\tan(28^\circ)$.

Step 2: Equate expressions for $h$ and solve for $x$ $$x \tan(42^\circ) = x \tan(28^\circ) + 50 \tan(28^\circ)$$ $$x[\tan(42^\circ) - \tan(28^\circ)] = 50 \tan(28^\circ)$$ $$x = \frac{50 \tan(28^\circ)}{\tan(42^\circ) - \tan(28^\circ)} \approx \frac{50(0.53171)}{0.90040 - 0.53171} = \frac{26.5855}{0.36869} \approx 72.109\text{ m}$$
Step 3: Calculate tower height $h$ $$h = 72.109 \times \tan(42^\circ) \approx 72.109 \times 0.90040 \approx \mathbf{64.9\text{ metres}}$$ Criterion A & D: Authentic survey modeling with verified algebraic elimination
Example 2 3D Angle Between Space Diagonal and Base Plane

Problem: A cuboid has dimensions $AB = 8\text{ cm}$, $BC = 6\text{ cm}$, and height $CG = 5\text{ cm}$. Calculate the angle that the space diagonal $AG$ makes with the base plane $ABCD$.

Step 1: Find the length of the base diagonal $AC$ $$AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}$$
Step 2: Identify the right triangle $\triangle ACG$

The projection of $AG$ onto the floor is $AC$. Line $CG$ is vertical, so $\angle ACG = 90^\circ$.

The desired angle is $\angle CAG = \theta$.

Step 3: Apply the tangent ratio $$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CG}{AC} = \frac{5}{10} = 0.5$$ $$\theta = \tan^{-1}(0.5) \approx \mathbf{26.6^\circ} \quad \text{(1 d.p.)}$$ Criterion C: Explicit spatial projection and precise trigonometric calculation

MYP 4–5 Practice Exercises: Trigonometric Ratios

Tier 1 (SOH CAH TOA skills), Tier 2 (Bearings & Inaccessible Heights), and Tier 3 (3D Triangles & Geometric Proofs).

Question 1 Tier 1: Standard 3 Marks

In a right-angled triangle, the hypotenuse is $14\text{ cm}$ and one acute angle is $35^\circ$. Find the length of the side opposite this angle to 3 significant figures.

View Full Method Markscheme
[M1] Set up the sine ratio: $\sin(35^\circ) = \frac{\text{Opp}}{14}$.
[M1] Rearrange: $\text{Opp} = 14 \sin(35^\circ)$.
[A1] Calculate: $\text{Opp} \approx 14 \times 0.573576 = \mathbf{8.03\text{ cm}}$.
Question 2 Tier 1: Standard 3 Marks

Find the measure of acute angle $\theta$ in a right-angled triangle where the adjacent leg is $11\text{ cm}$ and the hypotenuse is $15\text{ cm}$. Give your answer to 1 decimal place.

View Full Method Markscheme
[M1] Set up the cosine ratio: $\cos \theta = \frac{11}{15}$.
[M1] Apply inverse cosine: $\theta = \cos^{-1}\left(\frac{11}{15}\right)$.
[A1] Calculate: $\theta \approx \cos^{-1}(0.7333) = \mathbf{42.8^\circ}$.
Question 3 Tier 1: Standard 3 Marks

Evaluate the exact numerical value of the expression without a calculator: $4 \sin(30^\circ) \cos(60^\circ) + \tan(45^\circ)$.

View Full Method Markscheme
[M1] Substitute exact values: $\sin(30^\circ) = \frac{1}{2}$, $\cos(60^\circ) = \frac{1}{2}$, $\tan(45^\circ) = 1$.
[M1] Compute product: $4\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = 4 \times \frac{1}{4} = 1$.
[A1] Add: $1 + 1 = \mathbf{2}$.
Question 4 Tier 2: Intermediate 4 Marks

From the top of an $85\text{ m}$ high seaside cliff, the angle of depression to a sailing boat is $23^\circ$. How far is the boat from the base of the cliff?

View Full Method Markscheme
[M1] Angle of depression of $23^\circ$ equals the angle of elevation from the boat to the cliff top.
[M1] In right triangle: $\tan(23^\circ) = \frac{\text{Opp}}{\text{Adj}} = \frac{85}{d}$.
[M1] Solve for horizontal distance $d$: $d = \frac{85}{\tan(23^\circ)}$.
[A1] Calculate: $d \approx \frac{85}{0.424475} \approx \mathbf{200\text{ metres}}$ (3 s.f.).
Question 5 Tier 2: Intermediate 4 Marks

A ship sails $40\text{ km}$ on a bearing of $065^\circ$.
(a) How far north has the ship travelled?
(b) How far east has the ship travelled?

View Full Method Markscheme
[M1] Angle with North line is $65^\circ$. Distance is hypotenuse $40\text{ km}$.
[A1] North component: $\text{North} = 40 \cos(65^\circ) \approx 40(0.4226) = \mathbf{16.9\text{ km}}$.
[M1] East component: $\text{East} = 40 \sin(65^\circ)$.
[A1] East component: $40(0.9063) = \mathbf{36.3\text{ km}}$.
Question 6 Tier 2: Intermediate 4 Marks

Given that $\cos \theta = \dfrac{5}{13}$ where $\theta$ is an acute angle, find the exact values of:
(a) $\sin \theta$
(b) $\tan \theta$

View Full Method Markscheme
[M1] Draw a reference right triangle with adjacent $= 5$ and hypotenuse $= 13$.
[M1] Find opposite side by Pythagoras: $\text{Opp} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$.
[A1] Part (a): $\sin \theta = \frac{\text{Opp}}{\text{Hyp}} = \mathbf{\dfrac{12}{13}}$.
[A1] Part (b): $\tan \theta = \frac{\text{Opp}}{\text{Adj}} = \mathbf{\dfrac{12}{5}}$.
Question 7 Tier 2: Intermediate 4 Marks

A flagpole stands on level ground. From point $P$, the angle of elevation to the top of the pole is $38^\circ$. From point $Q$, which is $15\text{ m}$ directly behind $P$ in line with the pole, the angle of elevation is $26^\circ$. Find the height of the flagpole.

View Full Method Markscheme
[M1] Let height be $h$, distance from $P$ to base be $d$. Then $\tan(38^\circ) = \frac{h}{d} \implies d = \frac{h}{\tan(38^\circ)}$.
[M1] From $Q$: $\tan(26^\circ) = \frac{h}{d + 15} \implies d + 15 = \frac{h}{\tan(26^\circ)}$.
[M1] Substitute $d$: $\frac{h}{\tan(26^\circ)} - \frac{h}{\tan(38^\circ)} = 15 \implies h\left[\frac{1}{\tan(26^\circ)} - \frac{1}{\tan(38^\circ)}\right] = 15$.
[A1] Calculate: $\frac{1}{0.48773} - \frac{1}{0.78129} = 2.0503 - 1.2799 = 0.7704$. Then $h = \frac{15}{0.7704} \approx \mathbf{19.5\text{ metres}}$.
Question 8 Tier 3: Extended 5 Marks

In a regular right square-based pyramid $VABCD$, base $ABCD$ has side length $12\text{ cm}$ and all sloping edges are of length $10\text{ cm}$.
(a) Find the vertical height of the pyramid.
(b) Calculate the angle that the sloping edge $VA$ makes with the base plane $ABCD$.

View Full Method Markscheme
[M1] Find half-diagonal of base: full diagonal $AC = \sqrt{12^2 + 12^2} = 12\sqrt{2}$. Half-diagonal $AM = 6\sqrt{2}\text{ cm}$.
[M1] Right triangle $\triangle VMA$: $VM^2 + AM^2 = VA^2 \implies H^2 + (6\sqrt{2})^2 = 10^2$.
[A1] $H^2 + 72 = 100 \implies H^2 = 28 \implies \mathbf{H = \sqrt{28} = 2\sqrt{7} \approx 5.29\text{ cm}}$.
[M1] Angle $\angle VAM = \theta$: $\cos \theta = \frac{AM}{VA} = \frac{6\sqrt{2}}{10} = \frac{3\sqrt{2}}{5} \approx 0.8485$.
[A1] $\theta = \cos^{-1}(0.8485) \approx \mathbf{32.0^\circ}$.
Question 9 Tier 3: Extended 5 Marks

Prove the trigonometric identity algebraically for all acute angles $\theta$: $$\frac{1}{1 - \sin \theta} + \frac{1}{1 + \sin \theta} = 2 + 2\tan^2 \theta$$

View Full Method Markscheme
[M1] Combine fractions on LHS over common denominator: $$\text{LHS} = \frac{(1 + \sin \theta) + (1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)} = \frac{2}{1 - \sin^2 \theta}$$
[M1] Apply Pythagorean identity: $1 - \sin^2 \theta = \cos^2 \theta$. $$\text{LHS} = \frac{2}{\cos^2 \theta}$$
[M1] Write $2 = 2 \cdot \frac{\cos^2 \theta}{\cos^2 \theta}$: $$\frac{2}{\cos^2 \theta} = 2 \cdot \left(\frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta}\right) = 2\left(\frac{\sin^2 \theta}{\cos^2 \theta} + 1\right)$$
[A1] Recall $\frac{\sin \theta}{\cos \theta} = \tan \theta$, so $\frac{\sin^2 \theta}{\cos^2 \theta} = \tan^2 \theta$.
[R1] Thus $2(\tan^2 \theta + 1) = \mathbf{2 + 2\tan^2 \theta} = \text{RHS}$. [Proved]
Question 10 Tier 3: Extended 5 Marks

A camera drone at $D$ is $120\text{ m}$ above horizontal ground. From $D$, the pilot observes two cars $A$ and $B$ on the ground. Car $A$ is on a bearing of $040^\circ$ with angle of depression $35^\circ$. Car $B$ is on a bearing of $130^\circ$ with angle of depression $25^\circ$.
(a) Find the horizontal ground distance from the point directly beneath the drone to Car $A$ and Car $B$.
(b) Calculate the direct distance between Car $A$ and Car $B$.

View Full Method Markscheme
[M1] Let ground point beneath drone be $O$. Distance $OA = \frac{120}{\tan(35^\circ)} \approx \frac{120}{0.7002} \approx \mathbf{171.38\text{ m}}$.
[M1] Distance $OB = \frac{120}{\tan(25^\circ)} \approx \frac{120}{0.4663} \approx \mathbf{257.34\text{ m}}$.
[M1] Find angle between bearings: $\angle AOB = 130^\circ - 040^\circ = 90^\circ$!
[M1] Because $\angle AOB = 90^\circ$, apply Pythagoras to ground triangle $\triangle AOB$: $$AB = \sqrt{OA^2 + OB^2} = \sqrt{171.38^2 + 257.34^2}$$
[A1] $AB = \sqrt{29371 + 66224} = \sqrt{95595} \approx \mathbf{309\text{ metres}}$ (3 s.f.).

Trigonometric Ratios Reference

  • Sine: $\sin \theta = \dfrac{\text{Opp}}{\text{Hyp}}$
  • Cosine: $\cos \theta = \dfrac{\text{Adj}}{\text{Hyp}}$
  • Tangent: $\tan \theta = \dfrac{\text{Opp}}{\text{Adj}} = \dfrac{\sin \theta}{\cos \theta}$
  • Fundamental Identity: $\sin^2 \theta + \cos^2 \theta = 1$
  • Exact Values:
    • $\sin(30^\circ) = \cos(60^\circ) = \frac{1}{2}$
    • $\sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}$
    • $\sin(60^\circ) = \cos(30^\circ) = \frac{\sqrt{3}}{2}$
    • $\tan(30^\circ) = \frac{\sqrt{3}}{3}, \; \tan(45^\circ) = 1, \; \tan(60^\circ) = \sqrt{3}$
  • Bearings: Measured clockwise from North with $3$ digits.

Top Common MYP Pitfalls

1. Calculator in Radian Mode

Entering $\sin(35^\circ)$ on a calculator set to Radian mode instead of Degree mode. Always check the degree indicator!

2. Confusing Opposite and Adjacent

Failing to re-label sides when switching to the other acute angle in the triangle.

3. Misidentifying Angle of Depression

Measuring angle of depression from the vertical wall instead of the horizontal line of sight.

4. Bearings Counter-Clockwise

Measuring bearings counter-clockwise or from the East axis. Bearings are strictly clockwise from North!

Interactive Mini Test: Trigonometric Ratios

Answer the questions to see your score and feedback.

Question 1 of 5 Trigonometric Ratio Definition

In a right triangle with acute angle $\theta$, which ratio represents $\cos \theta$?

Question 2 of 5 Exact Value

What is the exact value of $\tan(60^\circ)$?

Question 3 of 5 Angle of Elevation and Depression

If the angle of depression from a lighthouse to a boat is $22^\circ$, what is the angle of elevation from the boat to the lighthouse?

Question 4 of 5 Back Bearing

If town $B$ is on a bearing of $075^\circ$ from town $A$, what is the bearing of town $A$ from town $B$?

Question 5 of 5 Pythagorean Identity

If $\sin \theta = 0.6$ for acute angle $\theta$, what is the value of $\cos \theta$?